OP Malhotra Class 11 Maths Solutions Chapter 17 Circle Chapter Test

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

S Chand Class 11 ICSE Maths Solutions Chapter 17 Circle Chapter Test

Question 1.
Find the centre and radius of the circle
2x² + 2y² – x = 0.
Solution:
Given eqn. of circle be
2x² + 2y² – x = 0 ⇒ x² + y² – \(\frac { x }{ 2 }\) = 0
∴ its centre be \(\left(\frac{1}{4}, 0\right)\)
and radius of circle = r = \(\sqrt{\left(\frac{1}{4}\right)^2+0}\) = \(\frac { 1 }{ 4 }\)

Question 2.
Find the equation of the circle with centre (- a, – b) and radius \(\sqrt{a^2-b^2}\).
Solution:
Using centre-radius form, eqn. of circle having centre (- a, – b) and radius \(\sqrt{a^2-b^2}\) is given by
(x + a)² + (y + b)² = (\(\sqrt{a^2-b^2}\))²
⇒ x² + y² + 2ax + 2by + a² + b² = a² – b²
⇒ x² + y² + 2ax + 2by + 2b² = 0

Question 3.
Find the equation of the circle drawn on the line joining (- 1, 2) and (3, – 4) as diameter.
Sol.
Using diameter form, eqn. of circle having the points (- 1, 2) and (3, – 4) as extremities of diameter is given by
(x + 1) (x – 3) + (y – 2) (y + 4) = 0
⇒ x² + y² – 2x + 2y – 11 = 0
which is the required eqn. of circle.

Question 4.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre lies on the line 4x + y = 16.
Solution:
Let the general eqn. of circle be given by
x² + y² + 2gx + 2fy + c = 0
Now circle (1) passes through the points (4, 1) and (6, 5).
16 + 1 + 8g + 2f + c = 0
⇒ 8g + 2f + c + 17 = 0 …(2)
36 + 25 + 12g + 10f + c = 0
⇒ 12g + 10f + c + 61 = 0
The centre (- g, – f) of circle (1) lies on line 4x + y = 16
∴ – 4g – f = 16
⇒ 4g + f + 16 = 0 …(3)
eqn. (3) – eqn. (2) gives;
4g + 8f + 44 = 0
⇒ g + 2f + 11 = 0
On solving eqn. (4) and eqn. (5); we have
7g + 21 = 0 ⇒ g = – 3 and f = – 4
∴ from (2) ;
– 24 – 8 + c + 17 = 0 ⇒ c = 15
putting the values of g, f and c in eqn. (1); we have
x² + y² – 6x – 8y + 15 = 0
which is the required eqn. of circle.

Question 5.
Find the equation of a circle of radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution:
Let the general eqn. of circle be
x² + y² + 2gx + 2fy + c = 0
Since centre of (1) i.e. (- g, – f) lies on x-axis.
∴ – f = 0 ⇒ f = 0
∴ eqn. (1) reduces to;
x² + y² + 2gx + c = 0 …(2)
Now circle given by eqn. (2) passes through the point (2, 3).
4 + 9 + 4g + c = 0
⇒ 4g + c + 13 = 0 …(3)
and radius of circle (2) = \(\sqrt{g²-c}\)
Also, given radius of circle be 5 .
∴ g² – c = 25 …(4)
On adding eqn. (3) and (4); we have
4g + g² – 12 = 0
⇒ (g – 2)(g + 6) = 0 ⇒ g = 2, – 6
When g = 2 ∴ from (3) c = – 21
Thus eqn. (2) reduces to;
x² + y² + 4x – 21 = 0
When g = – 6 ∴ from (3); c = 11
∴ eqn. (2) reduces to ;
x² + y² – 12x + 11 = 0
Thus eqns. (5) and (6) gives the required eqns. of circles.

Question 6.
Find the equation of the circle concentric with the circle x² + y² – 8x + 6y – 5 = 0 and passing through the point (- 2, – 7).
Solution:
eqn. of given circle be
x² + y² – 8x + 6y – 5 = 0
∴ Centre of circle (1) be (4, – 3)
and radius of circle = \(\sqrt{16+9+5}\) = \(\sqrt{30}\)
Since the required circle is concentric with circle (1).
∴ Centre of required circle be (4, – 3) and required circle passed through the point (- 2, – 7)
∴ radius of required circle
= \(\sqrt{(-2-4)^2+(-7+3)^2}\)
= \(\sqrt{36+16}\) = \(\sqrt{52}\)
Hence the required eqn. of circle having centre (4, – 3) and radius \(\sqrt{52}\) be given by
(x – 4)² + (y + 3)² = 52
⇒ x² + y² – 8x + 6y – 27 = 0

Question 7.
Find the equation of the circle through the points (0, 0), (2, 0) and (0, 4). Also find the coordinates of its centre and its radius.
Solution:
Let the general eqn. of circle be
x² + y² + 2gx + 2fy + c = 0
Circle (1) passes through the point (0, 0).
∴ 0 + 0 + 0 + 0 + c = 0 ⇒ c = 0
Further the points (2, 0) and (0, 4) also lies on eqn. (1).
∴ 4 + 0 + 4g = 0 ⇒ g = – 1
and 16 + 0 + 8f = 0 ⇒ f = – 2
putting the values of g, f and c in eqn. (1) ; we get
x² + y² – 2x – 4y = 0
which is the required eqn. of circle.
Further centre of circle be (- g, – f) i.e. (1, 2)
and radius of circle = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{1+4-0}\) = \(\sqrt{5}\)

Question 8.
Find the parametric representation of the circle x² + y² – 2x – 4y – 4 = 0.
Solution:
Given eqn. of circle be
x² + y² – 2x – 4y – 4 = 0
⇒ x² – 2x + y² – 4y – 4 = 0
⇒ (x² – 2x + 1) + (y² – 4y + 4) = 9
⇒ (x – 1)² + (y – 2)² = 9
⇒ \(\left(\frac{x-1}{3}\right)^2\) + \(\left(\frac{y-2}{3}\right)^2\) = 1 …(1)
Thus its parametric eqns. are ;
\(\frac { x – 1 }{ 3 }\) = cos θ ⇒ x = 1 + 3 cos θ
and \(\frac { y – 2 }{ 3 }\) = sin θ ⇒ y = 2 + 3 sin θ where θ be the parameter.

Question 9.
Find the length of the chord intercepted by the circle x² + y² = 25 on the line 2x – y + 5 = 0.
Solution:
Given eqn. of circle be x² + y² = 25 its centre be C(0, 0) and radius = 5
From C, drawn CM ⊥ chord AB Clearly it bisects the chord.

In right angled △AMC, we have
AC² = AM² + CM²
⇒ AM² = 5² – CM²
| CM | = length of ⊥ drawn from C(0, 0) to line 2x – y + 5 = 0
= \(\frac{|2 \times 0-0+5|}{\sqrt{2^2+(-1)^2}}\) = \(\frac{5}{\sqrt{5}}\) = \(\sqrt{5}\)
∴ from (1); AM² = 25 – 5
⇒ AM = \(\sqrt{20}\) = \(2 \sqrt{5}\)
∴ required length of chord intercepted by given circle on given line = AB = 2AM = 4\(\sqrt{5}\) units

Question 10.
Find the equations of the tangents to the circle x² + y² = 9, which are parallel to the line 3x + 4y = 0.
Solution:
Given eqn. of circle be
x² + y² = 9 …(1)
Its centre be C(0, 0) and radii 3 units and eqn. of given line be
3x + 4y = 0 …(2)
∴ eqn. of line parallel to line (2) be given by
3x + 4y + k = 0 …(3)
Since line (3) is tangent to circle (1).
∴ ⊥ distance drawn from C(0, 0) of circle to line (3) = radius of circle.
⇒ \(\frac{|3 \times 0+4 \times 0+k|}{\sqrt{3^2+4^2}}\) = 3 ⇒ \(\frac{|k|}{5}\) = 3
⇒ k = ± 15
∴ from (3) ; 3x + 4y ± 15 = 0
be the required equations of tangents to given circle.

Question 11.
Find the equation of the circle which touches the y-axis at a distance of +4 from the origin and cuts off an intercept 6 from the x-axis.
Solution:
There are two such circles one in first and other in second quadrant satisfying the given conditions.
Let C and D be the centres of both circles.
From C draw CM ⊥ PQ
∴ M be the mid-point of PQ.
MP = MQ = 3 units since PQ = 6 units

∴ In right angled △CPM, we have
CP² = CM² + PM² = 4² + 3² = 25
⇒ CP = 5 = radius of circle
⇒ CB = 5 units ⇒ OM = 5 units
∴ the coordinates of C and D are (5, 4) and (- 5, 4)
Thus required eqns. of circles are given by
(x ± 5)² + (y – 4)² = 5²
⇒ x² + y² ± 10 x – 8y + 16 = 0