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

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

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

Question 1.
Find the equations of the tangent to the circle 2x² + 2y² = 5 which are perpendicular to y = 2x.
Solution:
Given eqn. of circle be
2x² + 2y² = 5 ⇒ x² + y² = \(\frac { 5 }{ 2 }\) …(1)
centre of circle be (0, 0) and radius = \(\sqrt{\frac{5}{2}}\) and eqn. of given line be
y = 2x ⇒ 2x – y = 0 …(2)
Thus eqn. of line ⊥ to line (2) be
x + 2y + k = 0 …(3)
Now line (3) touches circle (1).
if length of ⊥ drawn from centre C(0, 0) of circle (1) to line (3) = radius of circle

∴ from (5); x + 2y ± \(\frac{5}{\sqrt{2}}\) = 0 be the

Question 2.
Find the equations of the tangents to the circle x² + y² – 8y – 8 = 0 which are parallel to the line 5x – 2y = 2.
Solution:
Given eqn. of circle be
x² + y² – 8y – 8 = 0
∴ Centre of circle (1) be (0, 4)
and radius of circle (1) = \(\sqrt{0+16+8}\)
= \(\sqrt{24}\) = \(2 \sqrt{6}\)
and eqn. of line parallel to given line
5x – 2y – 2 = 0
be given by 5x – 2y + k = 0 …(2)
Now line (2) touches circle (1).
if length of ⊥ drawn from centre (0, 4) of circle to line (2) = radius of circle .

putting the value of k in eqn. (2) ; we have 5x – 2y + 8 ± \(2 \sqrt{174}\) = 0 be the required eqns. of tangents to the given circle.

Question 3.
Find the equation of the circle which has extremities of a diameter the origin and the point (2, – 4). Find also the equations of the tangents to the circle which are parallel to this diameter.
Solution:
The eqn. of circle having extremities of a diameter as (0, 0) and point (2, – 4) is given by
(x – 0) (x – 2) + (y – 0) (y + 4) = 0
⇒ x² + y² – 2x + 4y = 0
eqn: of diameter using two point form, be given by
y – 0 = \(\frac{(-4-0)}{2-0}\)(x – 0)
⇒ y = – 2x ⇒ 2x + y = 0 …(2)
and eqn. of line parallel to eqn. (2) be given by
2x + y + k = 0 …(3)
Also, centre of circle be (1, – 2)
and radius of circle = \(\sqrt{1+4-0}\) = \(\sqrt{5}\)
Now line (3) is tangent to circle (1).
if length of ⊥ drawn from centre C(1, – 2) of circle to line (3) = radius of circle

∴ from (2); 2x + y ± 5 = 0 be the required eqns. of tangents to given circle.

Question 4.
Show that, whatever be the value of α, the lines x cos α + y sin α = a
and x sin α – y cos α = a
are tangents to the circle x² + y² = a². Hence obtain the locus of the points from which perperdicular tangents can be drawn to the circle x² + y² = a².
Solution:
Given eqn. of circle be
x² + y² = a² …(1)
∴ its centre be (0, 0) and radius a and given eqns. of lines are
x cos α + y sin α = a …(2)
and x sin α – y cos α = a …(3)
Here d₁ = ⊥ distance from centre C(0, 0) of circle (1) to line (2)

= \(\frac{a}{1}\) = a
∴ d₁ = d₂ = radius of given circle
Thus both lines (2) and (3) touch given circle (1) and hence are tangents to given circle.
Further slope of line (2) = m₁
= – \(\frac{\cos α}{\sin α}\) = – cot α
and slope of line (3) = m₂
= – \(\frac{\sin α}{-\cos α}\) = tan α
Here m₁ m₂ = (- cot α) tan α = – 1
Thus both tangents are perpendicular.
For finding the locus of a point from which two ⊥ tangents can be drawn, we have to eliminate θ from eqn. (2) and eqn. (3).
On squaring and adding eqn. (2) and eqn. (3); we have
(x cos α + y sin α)² + (x sin α – y sin α)²
= a² + a²
⇒ x² (cos² α + sin² α) + y²(sin ² α + cos² α) = 2a²
⇒ x² + y² = 2a² be the required locus.

Question 5.
Find the locus of the feet of the perpendiculars drawn from the point (b, 0) on tangents to the circle x² + y² = a².
Solution:
Given eqn. of circle be x² + y² = a² with centre (0, 0) and radius a.
eqn. of any tangent to given circle be
y = mx ± \(a \sqrt{1+m^2}\) …(1)
Let Q be the feet of ⊥ drawn from P(b, 0) on the tangent to the given circle.

∴ slope of line PQ be – \(\frac { 1 }{ m }\)
Thus eqn. of ⊥ PQ be given by
y – 0 = –\(\frac { 1 }{ m }\)(x – b)
⇒ my = – x + b
For locus of feet of ⊥, we eliminate m from eqn. (1) and eqn. (2); we have

On squaring both sides, we have
(y² +x² – bx)² = a² {y² + (b – x)²}
which is the required locus.

Question 6.
(i) Find the equation of that chord of the circle x² + y² = 15, which is bisected at the point (3, 2).
(ii) Find the locus of mid-points of all chords of the circle x² + y² = 15 that pass through the point (3, 4).
Solution:
(i) Given eqn. of circle be
x² + y² = 15 …(1)
its centre be C(0, 0) and radius = \(\sqrt{15}\)
Let M be the mid-point of chord AB.
∴ slope of CM = \(\frac { 2-0 }{ 3-0 }\) = \(\frac { 2 }{ 3 }\)
∴ slope of chord AB = – \(\frac { 3 }{ 2 }\) [∵ CM ⊥ AB]

Thus using one point form, eqn. of chord AB passes through M(3, 2) and having slope – \(\frac { 3 }{ 2 }\) be given by
y – 2 = –\(\frac { 3 }{ 2 }\)(x – 3)
⇒ 2y – 4 = – 3x + 9
⇒ 3x + 2y – 13 = 0

(ii) Given eqn. of circle be x² + y² = 15 its centre be (0, 0). Let M(α, β) be the midpoint of chord QR that passes through point P(3, 4).
slope of line CM = \(\frac{\beta-0}{\alpha-0}\) = \(\frac{\beta}{\alpha}\)
∴ slope of line QR = –\(\frac{\alpha}{\beta}\) [∵ QR ⊥ CM]

Thus eqn. of line QR be given by
y – β = –\(\frac{\alpha}{\beta}\)(x – α)
⇒ βy – β² = -αx + α²
and it passes through P(3, 4).
∴ 4β – β² = – 3α + α²
⇒ α² + β² – 3α – 4β = 0
Thus locus of M(α, β) be given by
x² + y² – 3x – 4y = 0

Question 7.
Find the locus of the middle points of the chords of the circle x² + y² = 4(y + 1) drawn through the origin.
Solution:
Given eqn. of circle be
x² + y² – 4y – 4 = 0 …(1)
∴ Centre of circle (1) be C (0, 2).
Let M(α, β) be the mid-point of chord QR that pass through O(0, 0).

Since CM ⊥ OQ
∴ \(\left(\frac{\beta-2}{\alpha}\right)\) \(\left(\frac{\beta}{\alpha}\right)\) = – 1
⇒ β² – 2β = – α²
⇒ α² + β² – 2β = 0
Thus required locus of M(α, β) be given by
x² + y² – 2y = 0