The equation of the circle which passes through the point (3,-2) and (-2,0) and centre line 2x-y=3,is

x^{2} + y^{2} - 3 x - 12 y + 2 = 0

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x^{2} + y^{2} - 3 x + 12 y + 2 = 0

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x^{2} + y^{2} + 3 x + 12 y + 2 = 0

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Solution

The correct option is D $x^{2} + y^{2} + 3 x + 12 y + 2 = 0$
Let $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ ........(1) be the equation $(3, - 2)$ lies on the circle.

Plugging in

$13 + 6 g - 4 f + c = 0$ .........(2)

$(- 2, 0)$ lies on the circle lies on the circle

$4 - 2 g + c = 0$ ........(3)

centre $(- g, - f)$ lies on $2 x - y - 3 = 0$

$- 2 g + f - 3 = 0$ ..........(4)

This (4) equation including the first and $3$ unknowns g,f, and c (Note $x$ and $y$ are considered constant) conditions for consistency given.

$∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + y^{2} & 2 x & 2 y & 1 13 & 6 & - 4 & 1 4 & - 4 x & 0 & 1 - 3 & - 2 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = 0$

After somehow operations and expanding (one can directly expand also)

$x^{2} + y^{2} + 3 x + 12 y + 2 = 0$

Suggest Corrections

Similar questions

2 x - y - 2 = 0

(3, - 2)

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The equation of the circle concentric with $x^{2} - 3 x + 4 y - c = 0$ and passing through (-1, -2) is

The equation of circle concentric with circle $x^{2} + y^{2} - 6 x + 12 y + 15$ = 0 and double its area is

x^{2} + y^{2} - 6 x + 12 y + 15 = 0

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