Students must practice this TS Intermediate Maths 2B Solutions Chapter 1 Circle Ex 1(d) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2B Solutions Chapter 1 Circle Exercise 1(d)

I.

Question 1.

Find the condition that the tangents are drawn from (0, 0) to S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0 be perpendicular to each other.

Solution:

If the angle between the tangents drawn from P(x_{1}, y_{1}) to S = 0 is θ then

Question 2.

Find the chord of contact of (0, 5) with respect to the circle x^{2} + y^{2} – 5x + 4y – 2 = 0.

Solution:

Comparing with x^{2} + y^{2} + 2gx + 2fy + c = 0,

we get g = \(-\frac{5}{2}\), f = 2, c = -2

∴ Equation of chord of contact of (x1, y1) w.r.t S = 0 is xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

∴ The equation of chord of contact of (0, 5) w.r.t. x^{2} + y^{2} – 5x + 4y – 2 = 0 is x(0) + y(5) + \(\left(\frac{-5}{2}\right)\) (x + 0) + 2(y + 5) – 2 = 0

⇒ 5y – \(\frac{5 x}{2}\) + 2y + 10 – 2 = 0

⇒ 5y – \(\frac{5 x}{2}\) + 2y + 8 = 0

⇒ 10y – 5x + 4y + 16 = 0

⇒ -5x + 14y + 16 = 0

⇒ 5x – 14y – 16 = 0

Question 3.

Find the chord of contact of (1, 1) to the circle x^{2} + y^{2} = 9.

Solution:

Equation of chord of contact of (x_{1}, y_{1}) w.r.t x^{2} + y^{2} = a^{2} is xx_{1} + yy_{1} – a^{2} = 0.

∴ The equation of chord of contact of (1, 1) w.r.t x^{2} + y^{2} = 9 is x(1) + y(1) – 9 = 0

⇒ x + y – 9 = 0

Question 4.

Find the polar of (1, 2) with respect to x^{2} + y^{2} = 7.

Solution:

Equation of polar of (x_{1}, y_{1}) with respect to x^{2} + y^{2} = a^{2} is xx_{1} + yy_{1} – 9 = 0.

∴ The equation of polar of (1, 2) with respect to x^{2} + y^{2} = 7 is x(1) + y(2) – 7 = 0

⇒ x + 2y – 7 = 0

Question 5.

Find the polar of (3, -1) with respect to the circle 2x^{2} + 2y^{2} = 11.

Solution:

Given equation of circle is x^{2} + y^{2} = \(\frac{11}{2}\)

∴ Equation of polar of (x_{1}, y_{1}) w.r.t x^{2} + y^{2} = a^{2} is xx_{1} + yy_{1} – a^{2} = 0.

⇒ x(3) + y(-1) – \(\frac{11}{2}\) = 0

⇒ 6x – 2y – 11 = 0

Question 6.

Find the polar of (1, -2) with respect to x^{2} + y^{2} – 10x – 10y + 25 = 0.

Solution:

Polar of (x_{1}, y_{1}) w.r.t x^{2} + y^{2} + 2gx + 2fy + c = 0 is xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

⇒ x(1) + y(-2) – 5(x + 1) – 5(y – 2) + 25 = 0

⇒ -4x – 7y + 30 = 0

⇒ 4x + 7y – 30 = 0

Question 7.

Find the pole of ax + by + c = 0 (c ≠ 0) w.r.t x^{2} + y^{2} = r^{2}

Solution:

Let (x_{1}, y_{1}) be the pole of ax + by + c = 0 ……(1) w.r.t. x^{2} + y^{2} = r^{2}; then the equation of polar is xx_{1} + yy_{1} – r^{2} = 0 ……….(2)

∴ From (1) and (2)

Question 8.

Find the pole of 3x + 4y – 45 = 0 w.r.t x^{2} + y^{2} – 6x – 8y + 5 = 0

Solution:

Let (x_{1}, y_{1}) be the pole of 3x + 4y – 45 = 0 ………(1)

Then the equation of polar of (x_{1}, y_{1}) w.r.t the circle x^{2} + y^{2} – 6x – 8y + 5 = 0 is xx_{1} + yy_{1} – 3(x + x1) – 4(y + y1) + 5 = 0

⇒ x(x_{1} – 3) + y(y_{1} – 4) – (3x_{1} + 4y_{1} – 5) = 0 ……….(2)

∴ From (1) and (2)

\(\frac{x_1-3}{3}=\frac{y_1-4}{4}=\frac{-\left(3 x_1+4 y_1-5\right)}{-45}\)

⇒ \(\frac{x_1-3}{3}=\frac{y_1-4}{4}\)

⇒ 4x_{1} – 12 = 3y_{1} – 12

⇒ 4x_{1} – 3y_{1} = 0 ……….(3)

\(\frac{y_1-4}{4}=\frac{3 x_1+4 y_1-5}{45}\)

⇒ 45y_{1} – 180 = 12x_{1} + 16y_{1} – 20

⇒ 12x_{1} – 29y_{1} + 160 = 0 ………(4)

Solving (3) and (4) we get

12x_{1} – 9y_{1} = 0

12x_{1} – 29y_{1} = -160

∴ 20y_{1} = 160

⇒ y_{1} = 8

and from (3)

4x_{1} = 24

⇒ x_{1} = 6

∴ Pole of 3x + 4y – 45 = 0 w.r.t x^{2} + y^{2} – 6x – 8y + 5 = 0 is (6, 8).

Question 9.

Find the pole of x – 2y + 22 = 0 w.r.t x^{2} + y^{2} – 5x + 8y + 6 = 0.

Solution:

Let (x_{1}, y_{1}) be the pole of the line x – 2y + 22 = 0 ………(1)

Then the equation of polar of (x_{1}, y_{1}) w.r.t circle x^{2} + y^{2} – 5x + 8y + 6 = 0 ………(2) is

⇒ 22y_{1} + 88 = 5x_{1} – 8y_{1} – 12

⇒ 5x_{1} – 30y_{1} – 100 = 0

⇒ x_{1} – 6y_{1} – 20 = 0 ………(5)

∴ 2x_{1} – 12y_{1} – 40 = 0 and 2x_{1} + y_{1} – 1 = 0

solving -13y_{1} – 39 = 0 ⇒ y_{1} = -3

and from (4), 2x_{1} – 3 – 1 = 0 ⇒ x_{1} = 2

pole of x – 2y + 22 = 0 w.r.t. x^{2} + y^{2} – 5x + 8y + 6 = 0 is (2, -3).

Question 10.

Show that the points (-6, 1) and (2, 3) are conjugate points w.r.t the circle x^{2} + y^{2} – 2x + 2y + 1 = 0.

Solution:

Let P(-6, 1) and Q(2, 3) be the given points.

Two points P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) one said to be conjugate

⇔ S_{12} = 0

⇔ x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = 0

∴ S_{12} = (-6)(2) + (1)(3) – 1(-6 + 2) + 1(1 + 3) + 1

= -12 + 3 + 4 + 4 + 1

= 0

∴ Hence the points P and Q are conjugate w.r.t. the given circle.

Question 11.

Show that the points (4, 2) and (3, -5) are conjugate points w.r.t the circle x^{2} + y^{2} – 3x – 5y + 1 = 0.

Solution:

Let P(x_{1}, y_{1}) = (4, 2) and Q(x_{2}, y_{2}) = (3, -5)

Now S_{12} = x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = 0

= (4)(3) + (2)(-5) – \(\frac{3}{2}\)(4 + 3) – \(\frac{5}{2}\)(2 – 5) + 1

= 12 – 10 – \(\frac{21}{2}+\frac{15}{2}\) + 1 = 0

= 24 – 20 – 21 + 15 + 2

= 0

Hence the points P and Q are conjugate w.r.t. the given circle.

Question 12.

Find the value of k if kx + 3y – 1 = 0, 2x + y + 5 = 0 are conjugate lines w.r.t the x^{2} + y^{2} – 2x – 4y – 4 = 0.

Solution:

Two lines l_{1}x + m_{1}y + n_{1} = 0 and l_{2}x + m_{2}y + n_{2} = 0 are said to be conjugate w.r.t. x^{2} + y^{2} + 2gx + 2fy + c = 0.

If r^{2}(l_{1}l_{2} + m_{1}m_{2}) = (l_{1}g + m_{1}f – n_{1}) (l_{2}g + m_{2}f – n_{2})

Here g = -1, f = -2,

r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{1+4+4}\) = 3

l_{1} = k, m_{1} = 3, n_{1} = -1, l_{2} = 2, m_{2}2 = 1, n_{2} = 5

∴ r^{2}(l_{1}l_{2} + m_{1}m_{2}) = 9(2k + 3)

and (l_{1}g + m_{1}f – n_{1}) (l_{2}g + m_{2}f – n_{2}) = (-k – 6 + 1) (-2 – 2 – 5) = (-k – 5) (-9)

∴ 9(2k + 3) = 9(k + 5)

⇒ 2k + 3 = k + 5

⇒ k – 2 = 0

⇒ k = 2

Question 13.

Find the value of k if x + y – 5 = 0, 2x + ky – 8 = 0 are conjugate w.r.t the circle x^{2} + y^{2} – 2x – 2y – 1 = 0.

Solution:

Comparing the given equation of circle with x^{2} + y^{2} + 2gx + 2fy + c = 0,

we have g = -1, f = -1, c = -1 and

radius = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{1+1+1}=\sqrt{3}\)

Condition for two lines l_{1}x + m_{1}y + n_{1} = 0 and l_{2}x + m_{2}y + n_{2} = 0 to be conjugate w.r.t. S = 0

r^{2} (l_{1}l_{2} + m_{1}m_{2}) = (l_{1}g + m_{1}f – n_{1}) (l_{2}g + m_{2}f – n_{2})

Given that the equations of lines are x + y – 5 = 0 and 2x + ky – 8 = 0

l_{1} = 1, m_{1} = 1, n_{1} = -5, l_{2} = 2, m_{2} = k, n_{2} = -8

∴ r^{2}(l_{1}l_{2} + m_{1}m_{2}) = 3(2 + k) and (l_{1}g + m_{1}f – n_{1}) (l_{2}g + m_{2}f – n_{2}) = (-1 – 1 + 5) (-2 – k + 8)

∴ 3(2 + k) = 3(6 – k) = 18 – 3k

⇒ 6 + 3k = 18 – 3k

⇒ 6k = 12

⇒ k = 2

Question 14.

Find the value of k if the points (1, 3) and (2, k) are conjugate w.r.t the circle x^{2} + y^{2} = 35.

Solution:

The condition for two points P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) to be conjugate w.r.t x^{2} + y^{2} = a^{2} is x_{1}x_{2} + y_{1}y_{2} – a^{2} = 0

Given P(x_{1}, y_{1}) = (1, 3) and Q(x_{2}, y_{2}) = (2, k) and a^{2} = 35.

∴ x_{1}x_{2} + y_{1}y_{2} – a^{2} = 2 + 3k – 35

⇒ 3k – 33 = 0

⇒ k = 11

Question 15.

Find the value of k if the points (4, 2), (k, -3) are conjugate points w.r.t the circle x^{2} + y^{2} – 5x + 8y + 6 = 0.

Solution:

Two points P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) are said to be conjugate w.r.t the circle S = 0 if S_{12} = 0.

i.e., x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = 0

Given P(x_{1}, y_{1}) = (4, 2) and Q(x_{2}, y_{2}) = (k, -3)

Then S_{12} = 4k – 6 – \(\frac{5}{2}\)(4 + k) + 4(2 – 3) + 6 = 0

⇒ 4k – 6 – \(\frac{5}{2}\)(4 + k) – 4 + 6 = 0

⇒ 4k – \(\frac{5}{2}\)(4 + k) – 4 = 0

⇒ 4k – 10 – \(\frac{5k}{2}\) – 4 = 0

⇒ 8k – 20 – 5k – 8 = 0

⇒ 3k – 28 = 0

⇒ k = \(\frac{28}{3}\)

II.

Question 1.

Find the angle between the tangents drawn from (3, 2) to the circle x^{2} + y^{2} – 6x + 4y – 2 = 0. (Mar. ’12)

Solution:

If θ is the acute angle between the tangents drawn from P(3, 2) to the circle S = 0 then

cos θ = \(\left|\frac{s_{11}-r^2}{s_{11}+r^2}\right|\) where r is the radius of the circle.

Where S_{11} = \(x_1^2+y_1^2\) + 2gx_{1} + 2fy_{1} + c

= 9 + 4 – 6(3) + 4(2) – 2

= 1

and r = radius of the given circle = \(\sqrt{9+4+2}=\sqrt{15}\)

∴ cos θ = \(\left|\frac{1-15}{1+15}\right|=\left|\frac{-14}{16}\right|=\frac{7}{8}\)

∴ θ = \({Cos}^{-1} \cdot\left(\frac{7}{8}\right)\)

∴ The angle between tangents drawn from (3, 2) to the given circle is \({Cos}^{-1} \cdot\left(\frac{7}{8}\right)\).

Question 2.

Find the angle between the pair of tangents drawn from (1, 3) to the circle x^{2} + y^{2} – 2x + 4y – 11 = 0.

Solution:

If θ is the acute angle between the tangents drawn from P(3, 2) to the circle S = 0 then

cos θ = \(\left|\frac{s_{11}-r^2}{s_{11}+r^2}\right|\) where r is the radius of the circle.

S_{11} = \(x_1^2+y_1^2\) + 2gx_{1} + 2fy_{1} + c

= 1 + 9 – 2 + 12 – 11

= 9

and r = radius of the given circle (r) = \(\sqrt{1+4+11}\) = 4

∴ cos θ = \(\left|\frac{9-16}{9+16}\right|=\left|\frac{7}{25}\right|\)

∴ θ = \({Cos}^{-1}\left(\frac{7}{25}\right)\)

∴ The angle between the pair of tangents drawn from (1, 3) to the circle x^{2} + y^{2} – 2x + 4y – 11 = 0 is \({Cos}^{-1}\left(\frac{7}{25}\right)\)

Question 3.

Find the angle between the pair of tangents drawn from (0, 0) to the circle x^{2} + y^{2} – 14x + 2y + 25 = 0.

Solution:

If θ is the angle between the tangents drawn from O(0, 0) to the circle S = 0 then

cos θ = \(\left|\frac{s_{11}-r^2}{s_{11}+r^2}\right|\)

where S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2\) + 2gx_{1} + 2fy_{1} + c

= 0 + 0 + 0 + 0 + 25

= 25

and r = \(\sqrt{49+1-25}\) = 5

∴ cos θ = \(\left|\frac{25-25}{25+5}\right|\) = 0

∴ θ = \(\frac{\pi}{2}\)

∴ The angle between the pair of tangents drawn from (0, 0) to the given circle is \(\frac{\pi}{2}\).

Question 4.

Find the locus of P if the tangents drawn from P to x^{2} + y^{2} = a^{2} include an angle α.

Solution:

The centre of the circle x^{2} + y^{2} = a^{2} is C = (0, 0) and radius = a

Question 5.

Find the locus of P if the tangents drawn from P to x^{2} + y^{2} = a^{2} are perpendicular to each other.

Solution:

Equation of the given circle is x^{2} + y^{2} = a^{2} and radius = a

Let P(x_{1}, y_{1}) be any point on the locus.

Since the tangents drawn from P(x_{1}, y_{1}) to the circle are perpendicular to each other,

the angle between tangents is \(\frac{\pi}{2}\)

Question 6.

Find the slope of the polar of (1, 3) with respect to the circle x^{2} + y^{2} – 4x – 4y – 4 = 0. Also, find the distance from the centre to it.

Solution:

Equation of polar of the point P(x_{1}, y_{1}) w.r.t circle S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0 is xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

∴ Polar of (1, 3) w.r.t. circle x^{2} + y^{2} – 4x – 4y – 4 = 0 is x(1) + y(3) – 2(x + 1) – 2(y + 3) – 4 = 0

⇒ -x + y – 12 = 0

⇒ x – y + 12 = 0

∴ The slope of the polar is ‘1’

Also, the distance from the centre (2, 2) to the polar of (1, 3) i.e., x – y + 12 = 0 is \(\left|\frac{2-2+12}{\sqrt{1+1}}\right|=\frac{12}{\sqrt{2}}=6 \sqrt{2}\)

Question 7.

If ax + by + c = 0 is a polar of (1, 1) w.r.t the circle x^{2} + y^{2} – 2x + 2y + 1 = 0 and H.C.F of a, b, c is equal to one then find a^{2} + b^{2} + c^{2}.

Solution:

Equation of polar of (1, 1) w.r.t the circle x^{2} + y^{2} – 2x + 2y + 1 = 0 is x(1) + y(1) – 1(x + 1) + 1(y + 1) + 1 = 0

⇒ 2y + 1 = 0 ……….(1)

Given that polar of P(1, 1) is ax + by + c = 0 ………(2)

since (1) and (2) represent the same line

\(\frac{\mathrm{a}}{0}=\frac{\mathrm{b}}{2}=\frac{\mathrm{c}}{1}=\mathrm{k}\)

⇒ a = 0, b = 2k, c = k

Given that H.C.F of a, b, c is ‘1’ we have k = 1

∴ a = 0, b = 2, c = 1

∴ a^{2} + b^{2} + c^{2} = 0 + 4 + 1 = 5

III.

Question 1.

Find the coordinates of the point of intersection of tangents at the points where x + 4y – 14 = 0 meets the circle x^{2} + y^{2} – 2x + 3y – 5 = 0.

Solution:

Let P(x_{1}, y_{1}) be the point of intersection of tangents drawn to the circle x^{2} + y^{2} – 2x + 3y – 5 = 0 …………(1) and given that x + 4y – 14 = 0 ………..(2) meets the circle at points A, B.

∴ Equation of chord of contact of (x_{1}, y_{1}) w.r.t circle (1) is xx_{1} + yy_{1} – (x + x_{1}) + \(\frac{3}{2}\)(y + y_{1}) – 5 = 0

⇒ \(x\left(x_1-1\right)+y\left(y_1+\frac{3}{2}\right)-\left(x_1-\frac{3}{2} y_1+5\right)=0\) …….(3)

Since (2) and (3) represent the same chord of contact

Question 2.

If the polar of the points on the circle x^{2} + y^{2} = a^{2} w.r.t the circle x^{2} + y^{2} = b2 touches the circle x^{2} + y^{2} = c^{2} then prove that a, b, c are in geometrical progression.

Solution:

Given equations of circles

x^{2} + y^{2} = a^{2} …….(1)

x^{2} + y^{2} = b^{2} …….(2)

x^{2} + y^{2} = c^{2} ……(3)

Let P(x_{1}, y_{1}) be any point on (1) then

\(x_1^2+y_1^2\) = a^{2} ………(4)

The equation of polar of P(x_{1}, y_{1}) w.r.t circle (2) is xx_{1} + yy_{1} – b^{2} = 0 ……..(5)

The Centre of (3) is (0, 0), and the radius = c.

Given that line (5) is polar of P w.r.t the circle (2) which touches the circle (3).

∴ The perpendicular distance from C(0, 0) of circle (3) to the line (5) = radius of circle (3)

∴ \(\frac{1-b^2 \text { । }}{\sqrt{x_1^2+y_1^2}}\) = c

⇒ \(\frac{\mathrm{b}^2}{\sqrt{\mathrm{a}^2}}\) = c

⇒ b^{2} = ac

⇒ a, b, c are in geometric progression.

Question 3.

Tangents are drawn to the circle x^{2} + y^{2} = 16 from the point P(3, 5). Find the area of the triangle formed by these tangents and the chord of contact of P.

Solution:

The given equation of a circle is x^{2} + y^{2} = 16

Centre C = (0, 0) and radius = 4

The equation of chord of contact of (3, 5) w.r.t. circle x^{2} + y^{2} = 16 is xx_{1} + yy_{1} – 16 = 0

⇒ 3x + 5y – 16 = 0

PL is the perpendicular distance from the point P(3, 5) to the chord AB.

∴ Area of the triangle formed by the tangents and the chord of P = \(\frac{108 \sqrt{2}}{17}\) square units.

Question 4.

Find the locus of the point whose polar with respect to the circles x^{2} + y^{2} – 4x – 4y – 8 = 0 and x^{2} + y^{2} – 2x + 6y – 2 = 0 are mutually perpendicular.

Solution:

Let P(x_{1}, y_{1}) be the locus of the point whose polars w.r.t the given circles x^{2} + y^{2} – 4x – 4y – 8 = 0 ……(1) and x^{2} + y^{2} – 2x + 6y – 2 = 0 ……(2) are mutually perpendicular.

Polar of P(x_{1}, y_{1}) w.r.t circle (2) is xx_{1} + yy_{1} – (x + x_{1}) + 3(y + y_{1}) – 2 = 0

⇒ x(x_{1} – 1) + y(y_{1} – 3) – (x_{1} – 3y_{1} + 2) = 0 ………(3)

Polar of P(x_{1}, y_{1}) w.r.t circle (1) is xx_{1} + yy_{1} – 2(x + x_{1}) – 2(y + y_{1}) – 8 = 0

⇒ x(x_{1} – 2) + y(y_{1} – 2) – (2x_{1} + 2y_{1} + 8) = 0 ………(4)

Slope of (3) is \(-\left(\frac{x_1-1}{y_1+3}\right)\)

and slope of (4) is \(\left(\frac{x_1-2}{y_1-2}\right)\)

If (3) and (4) are mutually perpendicular then

\(\left(\frac{x_1-1}{y_1+3}\right)\left(\frac{x_1-2}{y_1-2}\right)\) = -1

⇒ \(\left(x_1^2-3 x_1+2\right)=-\left(y_1^2+y_1-6\right)\)

⇒ \(x_1^2+y_1^2-3 x_1+y_1-4=0\)

∴ Locus of P(x_{1}, y_{1}) is x^{2} + y^{2} – 3x + y – 4 = 0.

Question 5.

Find the locus of the foot of the perpendicular drawn from the origin to any chord of the circle S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0 which subtends a right angle at the origin.

Solution:

Any chord of the circle may be taken as lx + my = 1 ………(1)

Given equation of circle is x^{2} + y^{2} + 2gx + 2fy + c = 0 ………(2)

Chord (1) subtends a right angle at the origin.

Hence the lines joining the origin to the points of intersection of (1) and (2) are obtained by homogenizing equation (2) with equation (1).

From (1), lx + my = 1

and from (2), x^{2} + y^{2} + (2gx + 2fy)(1) + c(1)^{2} = 0

⇒ x^{2} + y^{2} + (2gx + 2fy) (lx + my) + c(lx + my)^{2} = 0

⇒ x^{2} + y^{2} + (2glx^{2} + 2fy^{2}m + 2flm + 2gmxy) + c(l^{2}x^{2} + 2lmxy + m^{2}y^{2}) = 0

⇒ (1 + 2gl + cl^{2}) x^{2} + (1 + 2fm + cm^{2})y^{2} + xy(2fl + 2gm + 2lmc) = 0

This is a homogenizing second-degree equation that represents pair of lines OA and OB passing through the origin. If these lines are perpendicular then

the coefficient of x^{2} + coefficient of y^{2} = 0.

⇒ 1 + 2gl + 1 + cl^{2} + 2fm + cm^{2} = 0

⇒ c(l^{2} + m^{2}) + 2(gl + fm) + 2 = 0 ……….(3)

From (1), the equation of the line perpendicular to the chord is mx – ly = 0 ………..(4)

Solving lx + my = 1 …..(1) and mx – ly = 0 ………(4)

To find the locus of the foot of the perpendiculars we have to eliminate ‘l’ and ‘m’.

Substituting these values of l and m in (3), we get

⇒ c + 2(gx + fy) + 2(x2 + y2) = 0

⇒ 2(x^{2} + y^{2}) + 2(gx + fy) + c = 0

∴ Locus of the foot of the perpendiculars drawn from the origin to any chord of circle S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0 which subtends a right angle at the origin is 2(x^{2} + y^{2}) + 2(gx + fy) + c = 0.