Question

The centre of the circle, which cuts orthogonally each of the three circles $x^{2} + y^{2} + 2x + 17y + 4 = 0,$

$x^{2} + y^{2} + 7x + 6y + 11 = 0$ and $x^{2} + y^{2} - x + 22y + 3 = 0$ is

• A. (3, 2)
• B. N(1, 2)
• C. (2, 3)
• D. (0, 2)

Answer 1

Answer: (A)

(3, 2)

Answer 2

Let the circle is $x^{2} + y^{2} + 2gx + 2fy + c = 0$…..(i)

Circle (i) cuts orthogonally each of the given three circles. Then according to condition $2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}$

$2g + 17f = c + 4$…..(ii)

$7g + 6f = c + 11$…..(iii)

$- g + 22f = c + 3$…..(iv)

On solving (ii), (iii) and (iv), $g = - 3,f = - 2$.

Therefore, the centre of the circle $( - g, - f) = (3,2)$