Question

The equation of the image of the circle

$x^{2} + y^{2} + 16x - 24y + 183 = 0$ by the line mirror

$4x + 7y + 13 = 0$ is

• A. $x^{2} + y^{2} + 32x - 4y + 235 = 0$
• B. $x^{2} + y^{2} + 32x + 4y - 235 = 0$
• C. $x^{2} + y^{2} + 32x - 4y - 235 = 0$
• D. $x^{2} + y^{2} + 32x + 4y + 235 = 0$

Answer 1

Answer: (D)

$x^{2} + y^{2} + 32x + 4y + 235 = 0$

Answer 2

The given circle and line are

$x^{2} + y^{2} + 16x - 24y + 183 = 0$ …..(i) and

$4x + 7y + 13 = 0$ …..(ii)

Centre and radius of circle (i) are (– 8, 12) and 5 respectively. Let the centre of the image circle be $(x_{1},y_{1}).$

Then slope of $C_{1}C_{2} \times$ slope of $4x + 7y + 13 = - 1$

$\Rightarrow$ $\left( \frac{y_{1} - 12}{x_{1} + 8} \right) \times \left( - \frac{4}{7} \right) = - 1$ or $4y_{1} - 48 = 7x_{1} + 56$

or $7x_{1} - 4y_{1} + 104 = 0$…..(iii)

and mid point of $C_{1}C_{2}$ i.e., $\left( \frac{x_{1} - 8}{2},\frac{y_{1} + 12}{2} \right)$ lie on

$4x + 7y + 13 = 0$,

then $4\left( \frac{x_{1} - 8}{2} \right) + 7\left( \frac{y_{1} + 12}{2} \right) + 13 = 0$ or

$4x_{1} + 7y_{1} + 78 = 0$…..(iv)

Solving (iii) and (iv), we get $(x_{1},y_{1}) = ( - 16, - 2)$

$\therefore$ Equation of the image circle is $(x + 16)^{2} + (y + 2)^{2} = 5^{2}$ or $x^{2} + y^{2} + 32x + 4y + 235 = 0$