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

Students can cross-reference their work with Class 11 ISC Maths S Chand Solutions Chapter 17 Circle Ex 17(b) to ensure accuracy.

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

Question 1.
Find the centre and radius of the circle
(i) x² + y² + 4x – 4y – 1 = 0;
(ii) 2x² + 2y² = 3x – 5y + 7.
Solution:
(i) Given eqn. of circle be
x² + y² + 4x – 4y – 1 = 0
On comparing with
x² + y² + 2gx + 2fy + c = 0
we have, 2g = 4 ⇒ g = 2;
2f = – 4 c f = – 2 and c = – 1
∴ Centre of circle be (- g, – f) i.e. (- 2, 2).
and radius of circle = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{4+4-(-1)}\) = 3

(ii) Given eqn. of circle be
2x² + 2y² – 3x + 5y – 7 = 0
⇒ x² + y² – \(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 2 }\)y – \(\frac { 7 }{ 2 }\) = 0
On comparing with
x² + y² + 2gx + 2fy + c = 0
we have 2g = \(\frac { -3 }{ 2 }\)
⇒ g = \(\frac { -3 }{ 4 }\) and c = \(\frac { -7 }{ 2 }\)
2f = +\(\frac { 5 }{ 2 }\) ⇒ f = \(\frac { 5 }{ 4 }\)
∴ required centre of circle be(- g, – f) i.e. \(\left(\frac{3}{4}, \frac{-5}{4}\right)\)
and radius of circle = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{\frac{9}{16}+\frac{25}{16}+\frac{7}{2}}\)
= \(\sqrt{\frac{90}{16}}\)
= \(\frac{3 \sqrt{10}}{4}\)

Question 2.
Find the lengths of the intercepts of the circle 3x² + 3y² – 5x + 3y = 0 on the coordinates axes.
Solution:
Given eqn. of circle be,
3x² + 3y² – 5x + 3y = 0 …(1)
eqn. (1) meets x-axis i.e. y = 0, we have
3x² – 5x = 0 ⇒ x = 0, \(\frac { 5 }{ 3 }\)
∴ length of x-intercept = \(\frac { 5 }{ 3 }\)
For y-intercept, eqn. (1) meets y-axis at x = 0
∴ 3y² + 3y = 0 ⇒ y = 0, – 1
∴ length of y-intercept = 1

Question 3.
Find the equation of the circle, which passes through the point (5, 4) and is concentric with the circle
x² + y² – 8x – 12y + 15 = 0.
Solution:
eqn. of given circle be
x² + y² – 8x – 12y +15 = 0 …(1)
Centre of circle (1) be (4, 6).
Since required circle is concentric with circle (1).
∴ Centre of required circle be (4, 6)
and also required circle passes through the point (5, 4).
∴ radius of required circle
= \(\sqrt{(5-4)^2+(4-6)^2}\) = \(\sqrt{5}\)
Hence the required eqn. of circle having centre (4, 6) and radius \(\sqrt{5}\) be given by
(x – 4)² + (y – 6)² = 5
⇒ x² + y² – 8x – 12y + 47 = 0

Question 4.
The radius of the circle x² + y² – 2x + 3y + k = 0 is 2\(\frac { 1 }{ 2 }\). Find the value of k. Find also the equation of the diameter of the circle, which passes through the point \(\left(5,2 \frac{1}{2}\right)\).
Solution:
Given eqn. of circle be
x² + y² – 2x + 3y + k = 0 …(1)
On comparing with
x² + y² + 2gx + 2fy + c = 0
we have, 2g = – 2 ⇒ g = – 1;
2f =3 ⇒ f = \(\frac { 3 }{ 2 }\) and c = k
∴ radius of circle (1) = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{(-1)^2+\left(\frac{3}{2}\right)^2-k}\)
Also radius of circle (1) be 2\(\frac { 1 }{ 2 }\) i.e. \(\frac { 5 }{ 2 }\).
∴ \(\sqrt{1+\frac{9}{4}-k}\) = \(\frac { 5 }{ 2 }\) ⇒ \(\frac { 13 }{ 4 }\) – k = \(\frac { 25 }{ 4 }\)
⇒ k = \(\frac{13-25}{4}\) ⇒ k = – 3
Clearly centre of circle (1) be \(\left(1, \frac{-3}{2}\right)\)
Hence the required eqn. of diameter of circle which pass through the points \(\left(1, \frac{-3}{2}\right)\) and \(\left(5, \frac{5}{2}\right)\) be given by

Question 5.
Prove that the circle
x² + y² – 6x – 2y + 9 = 0
(i) touches the x-axis ;
(ii) lies entirely inside the circle
x² + y² = 18.
Solution:
(i) Equation of given circle be
x² + y² – 6x – 2y + 9 = 0 …(1)
it meets x-axis at y = 0
∴ from (1); we have
x² – 6x + 9 = 0 ⇒ (x – 3)² = 0 ⇒ x = 3
Thus given circle meets x-axis at one point (3, 0) i.e. circle (1) touches x-axis.

(ii) Centre of circle (1) be C₁(3, 1)
and radius of circle = \(\sqrt{3^2+1^2-9}\) = 1 = r₁
eqn. of given circle be
x² + y² = 18 …(2)

Centre of circle (2) be C₂(0, 0)
and radius of circle = r₂ = \(\sqrt{18}\) = \(3 \sqrt{2}\)
Now | C₁ C₂ | = \(\sqrt{(3-0)^2+(1-0)^2}\) = \(\sqrt{10}\) < \(\sqrt{18}\) = R
Also | C₁ C₂ | + r = \(\sqrt{10}\) + 1 < \(\sqrt{18}\) = R
Clearly the circle (1) lies entirely inside circle (2).

Question 6.
Find the co-ordinates of the centre of the circle x² + y² – 4x + 6y = 3. Given that the point A, outside the circle, has coordinates (a, b) where a and b are both positive, and that the tangents drawn from A to the circle are parallel to the two axes respectively, find the values of a and b.
Solution:
Given eqn. of circle be
x² + y² – 4x + 6y – 3 = 0 …(1)
On comparing eqn. (1) with
x² + y² + 2gx + 2fy + c= 0
we have 2g = -4 ⇒ g = – 2;
2f = 6 ⇒ f = 3 and c = – 3
∴ Centre of circle (1) be (- g, – f) i.e. (2, – 3)
and radius of circle (1) = \(\sqrt{(-2)^2+3^2+3}\) = 4

eqn. of line passing through point(a, b) and
having slope 0 (|| to x-axis) is given by
y – b = 0 …(1)
and eqn. of line || to y-axis be given by
x – a = 0 …(2)
both lines (1) and (2) are tangents.
∴ | – 3 – b | = radius of circle = 4
+ 3 + b = ± 4 ⇒ b = 1, – 7
and | 2 – a | = 4 ⇒ 2 – a = ± 4
⇒ a = 6, – 2
but a, b > 0
Thus a = 6 and b = 1

Question 7.
Find the equation of the circle whose centre is at the point (4, 5) and which touches the x-axis. Also find the coordinates of the points at which the circle cuts the y-axis.
Solution:
Since the circle with centre C(4, 5) touches x-axis.
∴ length of ⊥ from C(4, 5) on x-axis = radius of circle
⇒ | 5 | = radius of circle.
∴ required eqn. of circle is having centre (4, 5) and radius 5 be given by
(y – 4)² + (y – 5)² = 25
⇒ x² + y² – 8x – 10y + 16 = 0 …(1)

Now circle (1) meets y-axis at x = 0
∴ y² – 10y + 16 = 0
⇒ (y – 2)(y – 8) = 0
⇒ y = 2, 8
Thus the circle cuts y-axis at (0, 2) and (0, 8).

Question 8.
Prove that the circles
x² + y² – 4x + 6y + 8 = 0
and x² + y² – 10x – 6y + 14 = 0
touches at the point (3, – 1) .
Note. Apply the following geometric principles :
(i) Circles touch if d = r₁ ± r₂, where d is the distance between the centres and r₁, r₂ the radii of the circles.
(ii) The point of contact divides the central line in the ratio of their radii internally for external contact and externally for internal contact.
Here, C₁ ≡ (2, – 3)
and r₁ = \(\sqrt{4+9-8}\) = \(\sqrt{5}\);
C₂ ≡ (5, 3)
and r₂ = \(\sqrt{25+9-14}\) = \(2 \sqrt{5}\)
C₁C₂ = \(\sqrt{(2-5)^2+(-3-3)^2}\)
= \(\) = \(\) = r₁ + r₂
Hence, the given circles touch externally.
Now, the point of contact (x, y) divides C₁ C₂ in the ratio 1 : 2.

∴ The point of contact is (3, – 1).
Solution:
Given eqns. of circles are
x² + y² – 4x + 6x + 8 = 0 …(1)
x² + y² – 10x – 6y + 14 = 0 …(2)
∴ Centre C₁ of circle (1) be (2, – 3)
and r₁ = \(\sqrt{(-2)^2+3^2-8}=\sqrt{5}\)
Centre C₂ of circle (2) be (5, 3)
and r₂ = \(\sqrt{(-5)^2+(-3)^2-14}\)
= \(\sqrt{25+9-14}\) = \(\sqrt{20}\) = \(2 \sqrt{5}\)
∴ | C₁C₂ | = \(\sqrt{(5-2)^2+(3+3)^2}\) = \(\sqrt{9+36}\) = \(\sqrt{45}\) = \(3 \sqrt{5}\)
Thus | C₁C₂ | = r₁ + r₂
Therefore both circles touch each other externally at point P and P divides C₁C₂ in the ratio r₁ : r₂ i.e. 1 : 2 internally
∴ Coordinates of P are given by

Question 9.
Show that the circles x² + y² – 2x = 0 and x² + y² + 6x – 6y + 2 = 0 touch externally at the point \(\left(\frac{1}{5}, \frac{3}{5}\right)\).
Solution:
Given eqns. of circles are
x² + y² – 2x = 0 …(1)
and x² + y² +6x – 6y + 2 = 0 …(2)
Centre of circle (1) be C₁(+ 1, 0)
and radius of circle = r₁ = \(\sqrt{1^2+0^2}\) = 1
Centre of circle (2) be given by C₂(- 3, 3) and radius of circle = r₂
= \(\sqrt{(-3)^2+3^2-3}\)
= 4
Here | C₁C₂ | = \(\sqrt{(-3-1)^2+(3-0)^2}\)
= \(\sqrt{16+9}\) = 5
and r₁ + r₂ = 1 + 4 = 5
∴ | C₁ C₂ | = r₁ + r₂

Hence both circles touch each other externally.
The point P divides C₁ C₂ internally in the ratio r₁ : r₂ i.e. 1 : 4
Then coordinates of point P be given by \(\left(\frac{-3+4}{1+4}, \frac{3+0}{1+4}\right)\) i.e. \(\mathrm{P}\left(\frac{1}{5}, \frac{3}{5}\right)\)

Question 10.
Show that the circles x² + y² + 2x – 6y + 9 = 0 and x² + y² + 8x – 6y + 9 = 0 touch internally.
Solution:
Given eqns. of circles are
x² + y² + 2x – 6y + 9 = 0 …(1)
and x² + y² + 8x – 6y + 9 = 0 …(2)
Centre of circle (1) be C₁(- 1, 3)
and radius of circle (1) = r₁
= \(\sqrt{(1)^2+(-3)^2-9}\)
= 1
Centre of circle (2) be C₂(- 4, 3) and radius of circle (2)
= \(\sqrt{4^2+(-3)^2-9}=4\)

Here, | C₁ C₂ | = \(\sqrt{(-4+1)^2+(3-3)^2}\)
= \(\sqrt{9+0}\) = 3
and | r₁ – r₂ | = | 1 – 4 | = 3
∴ | C₁ C₂ | = | r₁ r₂ |
Thus both circles touch each other internally.

Question 11.
Find the equation of the circle which passes through the points (0, 0), (0, 1) and (2, 3).
Solution:
Let the general eqn. of circle be
x² + y² + 2gx + 2fy + c = 0 …(1)
Now circle (1) passes through (0, 0)
∴ 0 + 0 + 0 + 0 + c = 0 ⇒ c = 0
eqn. (1) passes through (0, 1)
∴ 0 + 1 + 0 + 2f + 0 = 0 ⇒ f = –\(\frac { 1 }{ 2 }\)
Further, circle (1) passes through the point (2, 3).
∴ 4 + 9 + 4g + 6f = 0
⇒ 13 + 4g – 3 = 0
⇒ g = \(\frac { -5 }{ 2 }\)
putting the values of g, f and c in eqn. (1) ;
we have x² + y² – 5x – y = 0 which is the required eqn. of circle.

Question 12.
Find the centre and radius of the circle which passes through the points (7, 5), (6, – 2), (- 1, – 1).
Solution:
Let the equation of circle be
x² + y² + 2gx + 2fy + c = 0 …(1)
Circle (1) passes through the point (7, 5).
∴ 49 + 25 + 14g + 10f + c = 0
⇒ 14g + 10f + c + 74 = 0 …(2)
The point (6, – 2) lies on eqn. (1).
∴ 36 + 4 + 12g – 4f + c = 0
⇒ 12g – 4f + c + 40 = 0 …(3)
Further circle (1) passes through the point (- 1, – 1).
∴ 1 + 1 – 2g – 2f + c = 0
⇒ – 2g – 2f + c + 2 = 0 …(4)
eqn. (2) – eqn. (3) gives ;
2g + 14f + 34 = 0
⇒ g + 7f + 17 = 0 …(5)
eqn. (3) – eqn. (4) gives ;
14g – 2f + 38 = 0
⇒ 7g – f + 19 = 0 …(6)
eqn. (5) + 7 × eqn. (6) gives ;
50g + 150 = 0 ⇒ g = – 3
∴ from (6); f = 7 × (- 3) + 19 = – 2
∴ from (2); – 42 – 20 + c + 74 = 0
⇒ c = – 12
putting the values of g, f and c in eqn. (1); we get
x² + y² – 6x – 4y – 12 = 0
which is the required eqn. of circle. Thus, Centre of circle be

and radius of circle = \(\sqrt{(-3)^2+(-2)^2+12}\) = 5

Question 13.
Find the equation of the circle circumscribing the triangle formed by the lines x + y + 1 = 0, 3x + y – 5 = 0 and 2x + y – 4 = 0.
Solution:
The eqns. of sides of triangle are
x + y + 1 = 0 …(1)
3x + y – 5 = 0 …(2)
2x + y – 4 = 0 …(3)
On solving eqn. (1) and eqn. (2) simultaneously
we have x = 3 and y = – 4 i.e. point of intersection of (1) and (2) be (3, – 4).
On solving eqn. (2) and eqn. (3) simultaneously we have x = 1 and y = 2.
On solving eqn. (1) and eqn. (3) simultaneously; we have x = 5 and y = -6
Thus the vertices of △ABC are A(3, – 4); B(1, 2) and C(5, – 6).
Let the eqn. of circumscribing circle be
x² + y² + 2gx + 2fy + c = 0 …(4)
point A(3, – 4) lies on eqn. (4).
9 + 16 + 6g – 8f + c = 0
⇒ 6g – 8f + c + 25 = 0 …(5)

Circle (5) passes through the point B(1, 2); we have
1 + 4 + 2g + 4f + c = 0
⇒ 2g + 4f + c + 5 = 0 …(6)
The point (5, – 6) lies on circle (4).
∴ 25 + 36 + 10g – 12f + c = 0
⇒ 10g – 12f + c + 61 = 0 …(7)
eqn. (5) – eqn. (6) gives ;
4g – 12f + 20 = 0
⇒ g – 3f + 5 = 0 …(8)
eqn. (7) – eqn. (6); we have
8g – 16f + 56 = 0
⇒ g – 2f + 7 = 0 …(9)
On solving eqn. (8) and eqn. (9); we get
f = – 2 and g = – 11
∴ from eqn. (6) ; we have
– 22 – 8 + c + 5 = 0 ⇒ c = 25
putting the values of g, f and c in eqn. (4);
we have x² + y² – 22x – 4y + 25 = 0
which is the required eqn. of circle.

Question 14.
Show that the circle x² + y² – 4x + 4y + 4 = 0 touches the co-ordinate axes. If the points of contact are A and B, find the equation of the circle which passes through A, B and the origin.
Solution:
Given eqn. of circle be
x² + y² – 4x – 4y + 4 = 0 …(1)
Circle (1) meets x-axis at y = 0
∴ from (1); x² – 4x + 4 = 0
⇒ (x – 2)² = 0 ⇒ x = 2
Thus circle (1) intersects x-axis at (2, 0) i.e. at one point of contact. ∴ Circle (1) touches x-axis.
Further circle (1) meets y-axis at x = 0
∴ from (1); y² + 4y + 4 = 0
⇒ (y + 2)² = 0 ⇒ y = – 2
Thus circle touches y-axis at (0, – 2).
Therefore given circle (1) touches both coordinate axes.
Thus the circle (1) meets the coordinate axes at A(2, 0) and B(0, – 2).
Let the general eqn. of circle be
x² + y² + 2gx + 2fy + c = 0 …(2)
Circle (1) passes through the point origin (0, 0), (2, 0) and (0, – 2).
∴ c = 0 ; 4 + 4g = 0 ⇒ g = – 1
and 4 – 4f = 0 ⇒ f = 1
putting the value of g, f and c in eqn. (2) ; we have x² + y² – 2x + 2y = 0 which is the required eqn. of circle.

Question 15.
Find the equation of the circle which passes through the points P(1, 0), Q(3, 0) and R(0, 2). Find also (i) the co-ordinates of the other point in which the axis of y cuts the circle, (ii) the coordinates of the other end of the diameter through Q.
Solution:
Let the general eqn. of circle be
x² + y² + 2gx + 2fy + c = 0 …(1)
Circle (1) passes through the point P(1, 0).
∴ 1 + 2g + c = 0 …(2)
The point Q(3, 0) lies on circle (1).
∴ 9 + 6g + c = 0 …(3)
Circle (1) passes through the point R(0, 2).
4 + 4 f + c = 0 …(4)
eqn. (3) – eqn. (2) gives ;
8 + 4g = 0 ⇒ g = – 2
∴ from (2) ; c = 3
∴ from (4) ; f = –\(\frac { 7 }{ 4 }\)
putting the values of g, f and c in eqn. (1) ; we have
x² + y² – 4x – \(\frac { 7 }{ 2 }\)y + 3 = 0
⇒ 2x² + 2y² – 8x – 7y + 6 = 0 …(5)
Circle (5) cutting y-axis at x = 0
∴ 2y² – 7y + 6 = 0
⇒ (y – 2) (2y – 3) = 0 ⇒ y = 2, \(\frac { 3 }{ 2 }\)
Thus, the circle (5) intersecting y-axis at (0, 2) and \(\left(0, \frac{3}{2}\right)\).
The centre of circle (5) be C \(\left(2, \frac{7}{4}\right)\).
Let the other end of diameter be Q’ (α, β). Then \(\left(2, \frac{7}{4}\right)\) be the mid-point of QQ’.

∴ \(\frac{3+\alpha}{2}\) = 2 ⇒ 3 + α = 4 ⇒ α = 1
and \(\frac{0+\beta}{2}\) = \(\frac { 7 }{ 4 }\) ⇒ β = \(\frac { 7 }{ 2 }\)
Thus, the coordinates of other end of diameter be \(\left(1, \frac{7}{2}\right)\).

Question 16.
Find the equation of the circle which has its centre on the line y = 2 and which passes through the points (2, 0) and (4, 0).
Solution:
Let the general eqn. of circle be given by
x² + y² + 2gx + 2fy + c = 0 …(1)
Circle (1) passes through the points (2, 0) and (4, 0).
∴ 4 + 4g + c = 0 …(2)
16 + 8g + c = 0 …(3)
eqn. (3) – eqn. (2) gives ;
4g + 12 = 0 ⇒ g = -3
∴ from (2) ; 4 – 12 + c = 0⇒ c = 8
Further centre (- g, – f) of circle (1) lies on the line y = 2
i.e. – f = 2 ⇒ f = – 2
putting the values of g, f and c in eqn. (1); we have
x² + y² – 6x – 4y + 8 = 0.
which is the required eqn. of circle.

Question 17.
Find the equation of the circle which passes through the points (1, – 2),(4, – 3) and has its centre on the line 3x + 4y + 10 = 0.
Solution:
Let the general eqn. of circle be
x² + y² + 2gx + 2fy + c = 0 …(1)
Now circle (1) passes through the point (1, – 2).
∴ 1 + 4 + 2g – 4f + c = 0
⇒ 2g – 4f + c + 5 = 0 …(2)
Further the point (4, – 3) lies on circle (1).
∴ 16 + 9 + 8g – 6f + c = 0
⇒ 8g – 6f + c + 25 = 0 …(3)
Further, centre (- g, – f) lies on the line
3x + 4y + 20 = 0
i.e. -3g -4f + 10 = 0
⇒ 3g + 4f – 10 = 0 …(4)
eqn. (3) – eqn. (2) gives ;
6g – 2f + 20 = 0
⇒ 3g – f + 10 = 0 …(5)
eqn. (4) – eqn. (5) gives ;
5f – 20 = 0 ⇒ f = 4
∴ from (5); 3g – 4 + 10 = 0 ⇒ g = – 2
∴ from (2); – 16 – 24 + c + 25 = 0 ⇒ c = 15
putting the values of g, f and c in eqn. (1) ;
we have x² + y² – 4x + 8y + 15 = 0
which is the required eqn. of circle.

Question 18.
The vertices A, B, C of a triangle ABC have co-ordinates (4, 4),(5, 3) and (6, 0) respectively. Find the equations of the perpendicular bisectors of AB and BC, the co-ordinates of the circumcentre and the radius of the circumcircle of the triangle ABC.
Solution:
Let D and E are the mid-points of sides ABand BC of △ABC
∴ Coordinates of D are \(\left(\frac{9}{2}, \frac{7}{2}\right)\) and E are \(\left(\frac{11}{2}, \frac{3}{2}\right)\)
slope of side AB = \(\frac{4-3}{4-5}\) = -1

∴ slope of line OD = \(\frac{-1}{-1}\) = 1 [∵ OD ⊥ AB]
Thus, required eqn. of OD is given by
y – \(\frac{7}{2}\) = 1 \(\left(x-\frac{9}{2}\right)\)
⇒ 2y – 7 = 1 \(\left(x-\frac{9}{2}\right)\)
⇒ 2y – 7 = 2x – 9
⇒ 2x – 2y – 2 = 0
⇒ x – y – 1 = 0 …(1)
Now slope of line BC = \(\frac{0-3}{6-5}\) = – 3
Since OE ⊥ BC
∴ slope of line OE = \(\frac { -1 }{ -3 }\) = \(\frac { 1 }{ 3 }\)
Thus, required eqn. of ⊥ bisector of BC be given by
y – \(\frac { 3 }{ 2 }\) = \(\frac { 1 }{ 3 }\)\(\left(x-\frac{11}{2}\right)\)
⇒ 6y – 9 = 2x – 11
⇒ 2x – 6y – 2 = 0
⇒ x – 3y – 1 = 0 …(2)
The point of intersection of these two ⊥ bisectors gives the centre of circumcircle.
On solving eqn. (1) and (2); we have
y = 0 ∴ x = 1
Thus (1, 0) gives the required circumcentre, and radius of circumcircle of △ABC
= | OB | =| OA | = | OC |
= \(\sqrt{(1-5)^2+(0-3)^2}\) = \(\sqrt{16+9}\) = 5

Question 19.
The radius of a circle is 5 units and it touches the circle x² + y² – 2x – 4y – 20 = 0 externally at the point (5, 5). Find the equation of the circle.
Solution:
Eqn. of given circle be
x² + y² – 2x – 4y – 20 = 0 …(1)
centre of circle (1) be C(1, 2)
and radius of circle = \(\sqrt{(-1)^2+(-2)^2+20}\) = 5 = r ₁
Let the coordinates of centre of required circle be C'(α, β).
Thus P(5, 5) divides the line segment CC’ in the ratio 1 : 1 internally

∴ 5 = \(\frac{1 \times 1+1 \times \alpha}{2}\) ⇒ α = 9
and 5 = \(\frac{2+\beta}{2}\) ⇒ β = 8
Thus centre of required circle be (9, 8).
Hence the required eqn. of circle having centre (9, 8) and radius 8 is given by
(x – 9)² + (y – 8)² = 5²
⇒ x² + y² – 18x – 16y + 120 = 0

Question 20.
A circle x² + y² – 2ax + 2y = 0 cuts the axis of x at the point A and it circumscribes an equilateral triangle with OA as one of the sides, where O is the origin. Find the value of a and the vertices of the equilateral triangle.
Solution:
eqn. of given circle be
x² + y² – 2ax + 2y = 0 …(1)
it meets x-axis at y = 0
∴ x² – 2ax = 0 ⇒ x = 0, 2a
Thus circle (1) intersects x-axis at O(0, 0) and A(2a, 0). Further, circle (1) meets y-axis at x =0.

i.e. at points (0, 0) and B(0, – 2)
Let OAC be the equilateral triangle inscribed in circle.
∠OBA = ∠OCA = 60° [angles lies in same segment]
∴ ∠OAB = 30°
In △OAB ; tan 60° = \(\frac { 2a }{ +2 }\) ⇒ a = \(\sqrt{3}\)
From C, draw CM ⊥ OA
∴ OM = \(\frac { OA }{ 2 }\) = \(\frac { 2a }{ 2 }\) = \(\sqrt{3}\)
and CM = a\(\sqrt{3}\) = 3
Thus the coordinates of equilateral triangle are O(0, 0); A(2\(\sqrt{3}\), 0) and C(\(\sqrt{3}\), – 3).

Question 21.
Find the equation of the circle which passes through the points (5, 0) and (1, 4) and whose centre lies on the line x + y – 3 = 0.
Solution:
Let the general eqn. of circle be
x² + y² + 2gx + 2fy + c = 0 …(1)
Circle (1) passes through the point (5, 0).
∴ 25 + 10g + c = 0 …(2)
The point (1, 4) lies on circle (1).
∴ 1 + 16 + 2g + 8f + c = 0
⇒ 2g + 8f + c + 17 = 0 …(3)
Further centre (- g, – f) of circle (1) lies on line x + y – 3 = 0
i.e. – g – f – 3 = 0 ⇒ g + f + 3 = 0 …(4)
eqn. (2) – eqn. (3) ; we have
8g – 8f + 8 = 0 ⇒ g – f + 1 = 0 …(5)
On solving eqn. (4) and eqn. (5); we have g = – 2 and f = – 1
∴ from (2); 25 – 20 + c = 0 ⇒ c = – 5
putting the values of g, f and c in eqn. (1); we have
x² + y² – 4x – 2y – 5 = 0
which is the required eqn. of circle.