Q94 / 160 · TS EAMCET 2021 (05 Aug Shift 1) · Mathematics

The equation of the circle passing through the points of intersection of two circles $x^{2} + y^{2} + 2 x + 3 y + 1 = 0, x^{2} + y^{2} + 4 x + 3 y + 2 = 0$ and the point $(- 1, 1)$ is

Solution

Using family of circle, required circle is given by $(S_{1} + λ S_{2} = 0)$

$(x^{2} + y^{2} + 2 x + 3 y + 1) + λ (x^{2} + y^{2} + 4 x + 3 y + 2) = 0$

It passes through $(- 1, 1)$

$\Rightarrow (1 + 1 - 2 + 3 + 1) + λ (1 + 1 - 4 + 3 + 2) = 0$

$\Rightarrow 4 + 3 λ = 0$

$\Rightarrow λ = \frac{- 4}{3}$

So, circle is,

$(3 x^{2} + 3 y^{2} + 6 x + 9 y + 3) - 4 x^{2} - 4 y^{2} - 16 x - 12 y - 8 = 0$

$\Rightarrow x^{2} + y^{2} + 10 x + 3 y + 5 = 0$

Hence, the correct answer is option: 'A'.