OP Malhotra Class 11 Maths Solutions Chapter 17 Circle Ex 17(c)

Access to comprehensive Class 11 ISC Maths S Chand Solutions Chapter 17 Circle Ex 17(c) encourages independent learning.

S Chand Class 11 ICSE Maths Solutions Chapter 17 Circle Ex 17(c)

Question 1.
The circle 4x² + 4y² = 25 cuts the line 3x + 4y – 10 = 0 at A and B. Calculate the coordinates of A and B.
Solution:
The eqn. of given line be
3x + 4y – 10 = 0 ⇒ y = \(\frac{10-3 x}{4}\) …(1)
Substituting the value of y in given circle
4x² + 4y² = 25
⇒ 4x² + 4\(\left(\frac{10-3 x}{4}\right)^2\) = 25
⇒ 4x² + \(\frac{(10-3 x)^2}{4}\) = 25
⇒ 16x² + 100 + 9x² – 60x = 100
⇒ 25x² – 60x = 0
⇒ 5x (5x – 12) = 0
⇒ x = 0, \(\frac{12}{5}\)
∴ from (1); y = \(\frac{10}{4}\), \(\frac{10-\frac{36}{5}}{4}\)
i.e. y = \(\frac{5}{2}\) ; \(\frac{7}{10}\)
Hence the required coordinates of A and B are \(\left(0, \frac{5}{2}\right)\) and \(\left(\frac{12}{5}, \frac{7}{10}\right)\).

Question 2.
Find the length of the chord x + 2y = 5 of the circle whose equation is x² + y² = 9. Determine also the equation of the circle described on the chord as diameter.
Solution:
given eqn. of line be x + 2y = 5 ⇒ \(y=\frac{5-x}{2}\)
putting the value of y in x² + y² = 9; we have
x² + \(\left(\frac{5-x}{2}\right)^2\) = 9 ⇒ 4x² + (5 – x)² = 36 ⇒ 5x² – 10x – 11 = 0

Question 3.
Find the intercept made by the circle 4x² + 4y² – 24x + 5y + 25 = 0 on the st. line 4x – 2y = 7.
Solution:
Given eqn. of circle be
4x² + 4y² – 24x + 5y + 25 = 0
⇒ x² + y² – 6x + \(\frac { 5 }{ 4 }\)y + \(\frac { 25 }{ 4 }\) = 0 …(1)
∴ Centre of circle (1) be \(\left(3, \frac{-5}{8}\right)\)

Question 4.
Find the equation of each circle satisfying the given conditions.
(i) Centre C(1, – 3) and tangent to 2x – y – 4 = 0.
(ii) Tangent to 2x – 3y – 7 = 0 at (2, – 1) and passes through (4, 1).
(iii) Tangent to 2x – 3y + 3 = 0 at (- 3, 6) and tangent to x + 3y – 7 = 0.
(iv) Tangent to 2x – 3y + 1 = 0 t(1, 1), radius \(\sqrt{13}\).
(v) Tangent to the y-axis at (0, \(\sqrt{13}\)) and passes through (- 1, 0).
Solution:
(i) Let r be the radius of required circle. eqn. of given tangent to circle be
2x – y – 4 = 0 …(1)

Thus the required eqn. of circle having centre C(1, – 3) and having radius \(\frac{1}{\sqrt{5}}\) is given by
(x – 1)² + (y + 3)² = \(\left(\frac{1}{\sqrt{5}}\right)^2\)
⇒ 5[x² + y² – 2x + 6y + 10] = 1
⇒ 5 (x² + y²) – 10x + 30y + 49 = 0

(ii) slope of line CP = \(\frac{\beta+1}{\alpha-2}\)
and slope of given line = \(\frac{-2}{-3}\) = \(\frac{2}{3}\)
Since radius of circle is ⊥ to te given tangent.

⇒ 2β + 2 = – 3α + 6
⇒ 3α + 2β – 4 = 0 …(1)
Also | CP | = | AC | = radius of circle
⇒ (α – 2)² + (β + 1)² = (α – 4)² + (β – 1)²
⇒ – 4α + 2β + 5 = – 8α – 2β + 17
⇒ 4α + 4β – 12 = 0
⇒ α + β – 3 = 0 …(2)
On solving eqn. (1) and eqn. (2) ; we have
α = – 2 and β = 5
Thus centre of required circle be (- 2, 5)
and radius of circle = \(\sqrt{(4+2)^2+(1-5)^2}\)
= \(\sqrt{52}\)
Hence the required eqn. of circle having centre (- 2, 5) and radius \(\sqrt{52}\) is given by
(x + 2)² + (y – 5)² = 52
⇒ x² + y² + 4x – 10y – 23 = 0

(iii) Let C(α, β) be the centre of required circle.
eqn. of given tangents are
3x + y + 3 = 0 …(1)
and x + 3y – 7 = 0 …(2)
slope of line CP = \(\frac{\beta-6}{\alpha+3}\)
and slope of line (1) = \(\frac{-3}{1}\) = – 3

Since radius is ⊥ to the tangent to required circle
∴ \(\left(\frac{\beta-6}{\alpha+3}\right)\) (-3) = -1
⇒ 3β – 18 = α + 3
⇒ α – 3β + 21 = 0 …(3)
Also length of ⊥ from C(α, β) to tangent (1) = length of ⊥ drawn from C(α, β) to tangent (2) = radius of circle
⇒ \(\frac{|3 \alpha+\beta+3|}{\sqrt{10}}\) = \(\frac{|\alpha+3 \beta-7|}{\sqrt{10}}\)
⇒ 3α + β + 3 = ± (α + 3β – 7)
Case-I : 3α + β + 3 = α + 3β – 7
⇒ 2α – 2β + 10 = 0 …(4)
On solving eqn. (3) and (4); we have
β = 8, α = 3
Thus radius of circle
= \(\sqrt{(3+3)^2+(8-6)^2}\) = \(\sqrt{40}\)
Hence required eqn. of circle having centre (3, 8) and radius \(\sqrt{40}\) is given by
(x – 3)² + (y – 8)² = 40
⇒ x² + y² – 6y – 16y + 33 = 0

Case-II.
When 3α + β + 3 = -α – 3β + 7
⇒ 4α + 4β – 4 = 0
⇒ α + β – 1 = 0 …(5)
On solving eqn. (3) and eqn. (5); we have
– 4β + 22 = 0 ⇒ β = \(\frac { 11 }{ 2 }\) and α = –\(\frac { 9 }{ 2 }\)
Thus required eqn. of circle having centre \(\left(-\frac{9}{2}, \frac{11}{2}\right)\) and radius \(\sqrt{\frac{5}{2}}\) is given by \(\left(x+\frac{9}{2}\right)^2\) + \(\left(y-\frac{11}{2}\right)^2\) = \(\frac{5}{2}\)
⇒ x² + y² + 9x – 11y + \(\frac{81+121}{4}\) – \(\frac{5}{2}\) = 0
⇒ x² + y² + 9x – 11y + 48 = 0
which is the required circle.

(iv) Let C(α, β) be the centre of circle.
slope of CP = \(\frac{\beta-1}{\alpha-1}\)
and slope of given tangent 2x – 3y + 1 = 0 = \(\frac{-2}{-3}\) = \(\frac{2}{3}\)

Since CP is ⊥ to given tangent
∴ \(\left(\frac{\beta-1}{\alpha-1}\right)\)\(\left(\frac{2}{3}\right)\) = -1
⇒ 2β – 2 = – 3α + 3
⇒ 3α + 2β – 5 = 0 …(1)
Since the line 2x – 3y + 1 = 0 is tangent to circle with centre (α, β).
∴ length of ⊥ from C(α, β) to given line = radius of circle
\(\frac{|2 \alpha-3 \beta+1|}{\sqrt{13}}\) = \(\sqrt{13}\)
⇒ 2α – 3β + 1 = ± 13
⇒ 2α – 3β – 12 = 0 …(2)
and 2α -3β + 14 = 0 …(3)
On solving eqn. (1) and simultaneously; we have
α = 3 ; β =- 2
On solving eqn. (1) and (3) simultaneously; we have
α = – 1 and β = 4
Thus required eqns. of circles are given by
(x – 3)² + (y + 2)² = 13
and (x + 1)² + (y – 4)² = 13

(v) Let C(α, β) be the centre of required circle and eqn. of y-axis be x = 0
Now x = 0 be the tangent to required circle
∴ CP ⊥ the line x = 0

and | CP | = | CA | ⇒ CP² = CA²
⇒ (α – 0)² + (β – \(\sqrt{13}\))² = (α + 1)² + (β – 0)²
⇒ α² + β² – 2\(\sqrt{13}\)β + 3 = α² + 2α +1 + β²
⇒ 2α + 2\(\sqrt{3}\)β – 2 = 0
α + \(\sqrt{3}\)β – 1 = 0 …(1)
putting β = \(\sqrt{3}\) in eqn. (1); we have
α + 3 – 1 = 0 ⇒ α = – 2
Thus (- 2, \(\sqrt{3}\)) be the centre of circle and radius of circle
= \(\sqrt{(-1+2)^2+(0-\sqrt{3})^2}\)
= \(\sqrt{1+3}\) = 2
Hence the required eqn. of circle having centre (- 2, \(\sqrt{3}\)) and radius 2 is given
by (x + 2)² + (y – \(\sqrt{3}\))² = 2²
⇒ x² + y² + 4x – 2\(\sqrt{3}\)y + 3 = 0

Question 5.
Find the length of the chord made by the axis of x, with the circle whose centre is (0, 3a) and which touches the st. line 3x + 4y = 37a.
Solution:
Given C(0, 3a) be the centre of circle and 3x + 4y = 37a be the tangent to circle whose centre C(0, 3a).
∴ length of ⊥ drawn from C(0, 3a) to given line 3x + 4y – 37a = 0 = radius of circle
⇒ radius of circle = \(\frac{|0+4 \times 3 a-37 a|}{5}\) = 5a

Thus the required eqn. of circle having centre (0, 3a) and radius 5a is given by
(x – 0)² + (y – 3a)² = 25a²
⇒ x² + y² – 6ay – 16a² = 0
it meets x-axis when y = 0
∴ from (1); x² – 16a² = 0 ⇒ x = ± 4a
∴ points of intersection of chord and circle
(1) are (0, 4a) and (0, – 4a).
∴ required length of chord
= \(\sqrt{(-4 a-4 a)^2}=|8 a|\)

Question 6.
Find the equation of the circle which has centre C(3, 1) and which touches the line 5x – 12y + 10 = 0.
Solution:
eqn. of given line be

Now line (1) is tangent to required circle whose centre C (3, 1).

∴ length of ⊥ drawn from C(3, 1) to line (1) = radius of circle = r
⇒ \(\frac{|5 \times 3-12 \times 1+10|}{\sqrt{5⇒+12⇒}}\) = radius of circle = r
⇒ r = \(\frac { 13 }{ 13 }\) = 1
Thus, the required eqn. of circle having centre C(3, 1) and radii 1 be given by
(x – 3)² + (y – 1)² = 1²
⇒ x² + y² – 6x – 2y + 9 = 0

Question 7.
Tangents from an external point. Find the equations of the tangents to the circle x² + y² = 10 through the external point (4, – 2).
Solution:
Given eqn. of circle be
x² + y² = 10 …(1)
Any tangent to circle (1) be given by
y = mx + \(\sqrt{10} \sqrt{1+m⇒}\) …(2)
[Here a =\(\sqrt{10}\)]
Now eqn. (2) passes through the point (4, – 2).
∴ – 2 = 4m + \(\sqrt{10} \sqrt{1+m⇒}\)
⇒ – 2 – 4m = \(\sqrt{10} \sqrt{1+m⇒}\)
On squaring both sides; we have
(2 + 4m)² = 10(1 + m²)
⇒ 16m² + 16m + 4 = 10 + 10 m²
⇒ 6m² + 16m – 6 = 0
⇒ 3m² + 8m – 3 = 0

Question 8.
Tangent satisfying given condition. Find the equations of the tangents to the circle.
(i) x² + y² = 25 inclined at an angle of 60° to the x-axis.
(ii) x² + y² + 2x + 2y = 7 inclined at an angle of 45° to the x-axis.
(iii) x²+ y² = a² making a triangle of area a² with the axes.
(iv) x²+ y² – 6x + 4y = 12 and parallel to the line 4x + 3y + 5 = 0.
(v) x² + y² – 22x – 4y + 25 = 0 and perp. to the line 5x + 12y + 9 = 0.
Solution:
(i) We know that y = mx + c be tangent to circle x² + y² = a² if c = ± \(a \sqrt{1+m²}\)
Here eqn. of given circle be
x² + y² = 25
∴ a² = 25 ⇒ a = 5
and m = tan 60° = \(\sqrt{3}\)
Thus required eqns. of tangent to circle be given by
\(y=\sqrt{3} x \pm 5 \sqrt{1+(\sqrt{3})^2}\)
\(\Rightarrow \quad y=\sqrt{3} x \pm 10\)

(ii) eqn. of given circle be
x² + y²+ 2x + 2y – 7 = 0
its centre C(- 1, – 1)
and radius of circle (1) = \(\sqrt{(1)^2+(1)^2+7}\) = 3
Here slope of tangent = m = tan 45° = 1
Let the eqn. of tangent to circle (1) be y = x + c …(2)

putting the value of c in eqn. (2) ; we have
x – y ± 3\(\sqrt{2}\) = 0 be the required equations of tangents to given circle.

(iii) eqn. of given circle be
x² + y² = a² …(1)
Let y = m x + c be the eqn. of tangent to circle (1) with centre (0, 0) and radius a.
∴ length of ⊥ from C(0, 0) to line (2) = radius of circle

eqn. of coordinate axes are x = 0 …(3)
and y = 0 …(4)
Now eqn. (2), (3) and (4) forms a triangle with area a².
Now eqn. (2) and (3) intersect at A(0, c) eqn. (3) and (4) intersect at B(0, 0)
and eqn. (2) and (4) intersect at \(\mathrm{C}\left(-\frac{c}{m}, 0\right)\)

Therefore required eqns. of tangents are
y = ± x ± \(\sqrt{2}\) a

(iv) eqn. of given line be
4x + 3y + 5 = 0 …(1)
Now eqn. of line parallel to line (1) be
4x + 3y + k = 0 …(2)
Now eqn. (2) is tangent to given circle
x² + y² – 6x + 4y – 12 = 0
and centre of circle be (3, – 2)
and r = \(\sqrt{(-3)^2+2²+12}\) = 5
∴ length of ⊥ drawn from C(3,- 2) to line (2) = radius of circle.

putting the value of k in eqn. (2) ; we have
4x + 3y + 19 = 0
and 4x + 3y – 31 = 0
are the required eqns. of tangents.

(v) eqn. of given line be
5x + 12y + 9 = 0 …(1)
∴ eqn. of line ⊥ to line (1) be
12x – 5y + k = 0 …(2)
and eqn. of given circle be
x² + y² – 22x – 4y + 25 = 0
its centre be (11, 2)
and r = radius of circle
= \(\sqrt{(-11)^2+(-2)^2-25}\) = 10
Now line (2) is tangent to circle (3) if length of ⊥ from C(11, 2) to line (2) = radius of circle .

putting these values of k in eqn. (2); we have 12x – 5y + 8 = 0
and 12x – 5y – 252 = 0
are the required eqns. of tangents to given circle (3).

Question 9.
Conditions of tangency. Find the condition that the circle x² + y² + 2gx + 2fy + c = 0 may touch
(i) the x-axis,
(ii) the y-axis, and
(iii) the x-axis, at the origin.
Solution:
eqn. of given circle be
x² + y² + 2gx + 2fy + c = 0 …(1)
centre of circle (1) be (-g, – f)
and radius of circle = \(\sqrt{g²+f²-c}\)

(i) eqn. of x-axis be y = 0
Now line (2) touches circle (1).
if ⊥ distance from centre C(- g, – f)
to line (2) = radius of circle

on squaring both sides, we have
f² = g² + f² – c
⇒ c = g² be the required condition.

(ii) eqn. of x-axis be x = 0 …(3)
Now line (3) touches circle (1).
if length of ⊥ drawn from centre (- g, – f)
to line (3) = radius of circle
⇒ \(\frac{|-g|}{1}\) = \(\sqrt{g²+f²-c}\)
⇒ g² =g² + f² – c ⇒ c = f²
which is the required condition.

(iii) Now line (2) tangent to circle (1).
if c = g² …(4)
Further circle (1) passes through (0, 0)
∴ c = 0
∴ g = 0 [using eqn. (4)]
Hence c = g = 0 be the required condition.

Question 10.
Find the conditions that the line
(i) y = mx + c may touch the circle x² + y² = a²
(ii) lx + my + n = 0 may touch the circle x² + y² + 2gx +2fy + c = 0.
Solution:
(i) Given eqn. of line be
y = mx + c ⇒ mx – y + c = 0 …(1)
and eqn. of given circle be
x² + y² = a² …(2)
Now centre of circle (2) be (0, 0) and radius a.
Since line (1) touches circle (2).
if length of ⊥ drawn from centre C(0, 0) of circle to line (1) = radius of circle.

which is the required condition.

(ii) given eqn. of line be
lx + my + n = 0 …(1)
and eqn. of circle be
x² + y² + 2gx + 2fy + c = 0
its centre be (- g, – f)
and radius = \(\sqrt{g^2+f^2-c}\)
Now line (1) touches circle (2)
if ⊥ distance from C(- g, – f) to line (1) = radius of circle .
\(\frac{|-l g-m f+n|}{\sqrt{l^2+m^2}}\) = \(\sqrt{g^2+f^2-c}\)
On squaring both sides; we have
(lg + mf – n)² = (l² + m²) (g² + f² – c)
which is the required condition.

Question 11.
For what value of k will the line 4x + 3y + k = 0 touch the circle 2x² + 2y² = 5x?
Solution:
The eqn. of given line be
4x + 3y + k = 0 …(1)
given eqn. of circle be
x² + y² – \(\frac { 5x }{ 2 }\) = 0 …(2)
∴ Centre of circle be \(\left(\frac{5}{4}, 0\right)\)
and radius of circle = \(\sqrt{\left(\frac{5}{4}\right)^2+0^2-0}\) = \(\frac { 5 }{ 4 }\)
Since line (1) touches circle (2)
if length of ⊥ drawn from centre C\(\left(\frac{5}{4}, 0\right)\) of circle to line (1) = radius of circle

Question 12.
Show that 3x – 4y + 11 = 0 is a tangent to the circle x² + y² – 8y + 15 = 0 and find the equation of the other tangent which is parallel to the st. line 3x = 4y.
Solution:
Given eqn. of line be
3x – 4y + 11 = 0 …(1)
and eq. of circle be
x² + y² – 8y + 15 = 0 …(2)
its centre be (0, 4)
and radius of circle = \(\sqrt{0^2+16-15}\)
= 1 = r
Now d = length of ⊥ drawn from centre (0, 4) of circle to line (1)
= \(\frac{|3 \times 0-4 \times 4+11|}{\sqrt{3^2+(-4)^2}}\) = \(\frac{5}{5}\) = 1
∴ d = r
Thus line (1) touches circle (2).
eqn. of line parallel to line 3x = 4y be given by 3x – 4y + k = 0 …(3)
Now line (3) is tangent to circle (2)
if length of ⊥ drawn from C(0, 4) to line (3) = radius of circle
if \(\frac{|3 \times 0-4 \times 4+k|}{\sqrt{3^2+(-4)^2}}=1\)
⇒ | k – 16 | = 5 ⇒ k = ± 5 + 16
⇒ k = 21, 11
Thus the required eqn. of other tangent be 3x – 4y + 21 = 0

Question 13.
Show that x = 7 and y =8 touch the circle x² + y² – 4x – 6y – 12 = 0 and find the points of contact.
Solution:
eqn. of given circle be
x² + y² – 4x – 6y – 12 = 0 …(1)
putting x = 7 in eqn. (1); we have
7² + y² – 28 – 6y – 12 = 0
⇒ y² – 6y + 9 = 0
⇒ (y – 3)² = 0 …(2)
Thus eqn. (2) which is quadratic in x have equal roots.
∴ line x = 7 touches circle (1).
Thus y = 3 ∴ (7, 3) be the required point of contact.
Further when y = 8; eqn. (1) reduces to ;
x² + 64 – 4x – 48 – 12 = 0
⇒ (x – 2)² = 0 …(3)
⇒ x = 2
Thus eqn. (3) has two equal roots.
Therefore line y = 8 touches circle (1) and have one point of contact given by (2, 8).

Question 14.
Show that the line 3x + 4y + 20 = 0 touches the circle x² + y² = 16 and find the point of contact.
Solution:
Given eqn. of line be
3x + 4y + 20 = 0 …(1)
and given eqn. of circle be
x² + y² = 16 …(2)
From (1); y = \(\frac{-20-3 x}{4}\)
putting the value of y in eqn. (2); we have
x² + \(\left(\frac{20+3 x}{4}\right)^2\) = 16
⇒ 16x² + (20 + 3x)² = 256
⇒ 25x² + 120x + 144 = 0
⇒ (5x + 12)² = 0 …(3)
Thus eqn. (3) which is a quadratic in x have equal roots.
∴ line (1) touches circle (2).
∴ x = –\(\frac{12}{5}\) ∴ from (1); we have
\(\frac{-36}{5}\) + 4y + 20 = 0
⇒ \(\frac{64}{5}\) + 4y = 0 ⇒ y = \(\frac{-16}{5}\)
Hence the quired point of contact be \(\left(\frac{-12}{5}, \frac{-16}{5}\right)\)

Question 15.
Length of the tangent. Prove that the length t of the tangent from the point P(x₁, y₁) to the circle x² + y² + 2gx + 2fy + c = 0 is given by t = \(\sqrt{x_1^2+y_1^2+2 g x_1+2 f y_1+c}\)
Hence, find the length of the tangent
(i) to the circle x² + y² – 2x – 3y – 1 = 0 from the point (2, 5);
(ii) to the circle x² + y² – 6x + 8y + 4 = 0 from the origin ;
(iii) to the circle 3x² + 3y² -7x – 6y = 12 from the point (6, – 7);
(iv) to the circle x² + y² – 4y – 5 = 0 from the point (4, 5).
Solution:
Given eqn. of circle be
x² + y² + 2gx + 2fy + c = 0 …(1)
Centre of (1) be C(- g,- f)
and radius of circle = CQ = \(\sqrt{g^2+f^2-c}\)

In right angled △CQP, we have
CP² = CQ² + t²
⇒ t² = CP² – CQ²
= (x₁ + g)² + (y₁)² – (g² + f² – c)
= x₁² + y₁² + 2gx₁ + 2fy₁ + c
⇒ t = \(\sqrt{x_1^2+y_1^2+2 g x_1+2 f y_1+c}\)

(i) Required length of tangent to given circle from point (2, 5)
= \(\sqrt{2^2+5^2-2 \times 2-3 \times 5-1}\)
= \(\sqrt{4+25-4-15-1}\) = 3

(ii) Required length of the tangent to given circle from the origin
= \(\sqrt{0^2+0^2-6 \times 0+8 \times 0+4}\) = 2

(iii) Given eqn. of circle can be written as
x² + y² – \(\frac { 7 }{ 3 }\)x – 2y – 4 = 0
∴ required length of tangent to given circle from point (6, – 7)
= \(\sqrt{6^2+(-7)^2-\frac{7}{3} \times 6-2 \times(-7)-4}\)
= \(\sqrt{36+49-14+14-4}\) = \(\sqrt{81}\) = 9

(iv) Required length of tangent to given circle from the point (4, 5)
= \(\sqrt{4^2+5^2-4 \times 5-5}\) = 4

Question 16.
If x = 4 + 5 cos θ and y = 3 + 5 sin θ, show that the locus of the point (x, y) as θ varies, is a circle. Find the centre and radius of the circle.
Solution:
Given x = 4 + 5 cos θ
⇒ x – 4 = 5 cos θ …(1)
and y = 3 + 5 sin θ
⇒ y – 3 = 5 sin θ …(2)
To find locus of point (x, y), we eliminate θ from eqn. (1) and eqn. (2).
On squaring and adding (1) and (2); we have
(x – 4)² + (y – 3)² = 25(cos² θ + sin² θ)
⇒ (x – 4)² + (y – 3)² = 25
which is clearly represents a circle with centre (4, 3) and radius 5 .

Question 17.
A(1, 0) and B(7, 0) are two points on the axis of x. A point P is taken in the first quadrant such that PAB is an isosceles triangle and PB=5 units. Find the equation of the circle described on PA as diameter.
Solution:
Let the coordinates of P are P(α, β).

Since △PAB be an isosceles triangle
∴ PA = PB ⇒ PA² = PB²
⇒ (α – 1)² + (β – 0)² = (α – 7)² + (β – 0)²
⇒ α² – 2a + 1 + β² = α² – 14α + 49 + β²
⇒ 12α = 48 ⇒ α = 4
Also | PB | = 5 units ⇒ PB² = 25
⇒ (α – 7)² + (β – 0)² = 25
⇒ (4 – 7)² + β² = 25
⇒ β² = 25 – 9 = 16
⇒ β = ± 4
but point P lies in first quadrant
∴ coordinates of P are (4, 4).
Hence the required eqn. of circle having PA as diameter be given by
(x – 4) (x – 1) + (y – 4)(y – 0) = 0
⇒ x² + y² – 5x – 4y + 4 = 0

Question 18.
Find the equation of the circle which touches the line y = 2, passes through the origin and the point where the curve y² – 2x + 8 = 0 meets the x-axis.
Solution:
The eqn. of curve be
y² – 2x + 8 = 0 …(1)
Now curve (1) meets x-axis when y = 0
∴ 2x = 8 ⇒ x = 4
∴ Curve (1) meets x-axis at (4, 0).
Let (h, k) be the centre of required circle and r be the required of required circle.
Thus the required eqn. of circle be
(x – h)² + (y – k)² = r² …(2)
eqn. (2) passes through (0, 0).
∴ h² + k² = r² …(3)
Further eqn. (2) passes through the point (4, 0).
∴ (4 – h)² + k² = r² …(4)
Also the circle (2) touches line
y – 2 = 0 …(5)
∴ ⊥ distance from C(h, k) to line (5) = radius of circle
\(\frac{|k-2|}{1}=r\) …(6)
From (3) and (4); we have
h² + k² = (4 – h)² + k²
h² = h² – 8h + 16 ⇒ h = 2
Also from eqn. (3) and (6); we have
2² + k² = (k – 2)²
⇒ 4 + k² = k² + 4 – 4k ⇒ k = 0
∴ from (6) ; r = | 0 – 2 | = 2
putting the values of h, k and r in eqn. (2);
We have
(x – 2)² + (y – 0)² = 2²
⇒ x² + y² – 4x = 0
which is the required eqn. of circle.

Question 19.
(i) Prove that the line y = 2x touches the circle x² + y² + 16x + 12y + 80 = 0 and find the co-ordinates of the point of contact.
(ii) The circle x² + y² – 6x – 10y + p = 0 does not intersect or touch either axis and the point (1, 4) is inside the circle. Calculate the range of possible values of p.
Solution:
(i) Given eqn. of line be y = 2x …(1)
and eqn. of given circle be
x² + y² + 16x + 12y + 80 = 0 …(2)
Substituting the value of y from eqn. (1) in eqn. (2); we have
x² + 4x² + 16x + 24x + 80 = 0
⇒ 5x² + 40x + 80 = 0
⇒ x² + 8x + 16 = 0
⇒ (x + 4)² = 0 …(3)
eqn. (3) is a quadratic in x and have equal roots.
∴ line (1) touches circle (2).
∴ from (3) ; x = – 4
and from (1); y = – 8
Thus the required coordinates of point of contact be (- 4, – 8).

(ii) Given eqn. of circle be
x² + y² – 6x – 10y + p = 0 …(1)
Its centre be C(3, 5)
and radius of circle (1) = r
= \(\sqrt{9+25-p}\) = \(\sqrt{34-p}\)
and also given point be A (1, 4).
Now | CA | = \(\sqrt{(3-1)^2+(5-4)^2}\)
= \(\sqrt{4+1}\) = \(\sqrt{5}\)
Since the point A (1, 4) lies inside the circle.
∴ | CA | < r
\(\sqrt{5}\) < \(\sqrt{34-p}\)
⇒ 5 < 34 – p
⇒ p < 29 …(2) Further circle (1) does not intersect or touch coordinate axes. ∴ d > r [For y-axis]
⇒ 3 > \(\sqrt{34-p}\)
⇒ 9 > 34 – p
⇒ p > 25 …(3)

For x-axis ; y = 0 ∴ 5 > \(\sqrt{34-p}\)
⇒ 25 > 34 – p ⇒ p > 9 …(4)
∴ from eqn. (3) and (4) ; we have
p > 25 …(5)
From eqn. (2) and eqn. (5) ; we have
25 < p < 29