Matrix Matching : $(2, - 3)$

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The centre of the circle $x^{2} + y^{2} - 4 x + 6 y - 11 = 0$

The radius of the circle $x^{2} + y^{2} - 4 x - 2 y - 4 = 0$

The equation of the circle whose ends of diameter $(2, - 1)$ and $(- 3, 4)$

If one end of the diameter of circle $x^{2} + y^{2} - 5 x - y + 4 = 0$ is $(2, - 1)$ then the other end is

$(3, 2)$

$x^{2} + y^{2} + x - 3 y - 10 = 0$

$(2, - 3)$

answer is D, C, B, A.

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Detailed Solution

A) $x^{2} + y^{2} - 4 x + 6 y - 11 = 0$

Centre $(- g, - f) = (2, - 3)$ .

B) Radius $= \sqrt{g^{2} + f^{2} - c}$

$\begin{array}{l} x^{2} + y^{2} - 4 x - 2 y - 4 = 0 \\ g = - 2, f = - 1, c = - 4 \\ = \sqrt{{(- 2)}^{2} + {(- 1)}^{2} - (- 4)} \\ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \end{array}$

Question Image

Given ends of diameter $(2, - 1), (- 3, 4)$

Centre of the circle $(\frac{2 - 3}{2}, \frac{- 1 + 4}{2})$

Radius $= \sqrt{{(2 + \frac{1}{2})}^{2} + {(- 1 - \frac{3}{2})}^{2}}$

$= \sqrt{\frac{25}{4} + \frac{25}{4}} = \sqrt{\frac{2 (25)}{4}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}}$

$∴$ Equation of the circle ${(x + \frac{1}{2})}^{2} + {(y - \frac{3}{2})}^{2} = {(\sqrt{\frac{25}{2}})}^{2}$

$\begin{array}{l} x^{2} + \frac{1}{4} + x + y^{2} + \frac{9}{4} - 3 y = \frac{25}{2} \\ x^{2} + y^{2} + x - 3 y = \frac{25}{2} - \frac{5}{2} \\ x^{2} + y^{2} + x - 3 y = 10 \\ ∴ x^{2} + y^{2} + x - 3 y - 10 = 0 \end{array}$