Q93 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics
The centre of the circle passing through the points of intersection of the circles $(x+3)^2+(y+2)^2=25$ and $(x-2)^2+(y-3)^2=25$ and cutting the circle $(x+1)^2+(y-2)^2=16$ orthogonally is
- Correct answerA$\left(\frac{-27}{2}, \frac{-25}{2}\right)$
- B$(0,0)$
- C$\left(\frac{16}{3}, \frac{-25}{4}\right)$
- D$\left(\frac{4}{7}, \frac{3}{7}\right)$
Solution
Equation of circle passes through the point of intersections of given circles
$S_1 \equiv(x+3)^2+(y+2)^2=25$
and $\quad S_2=(x-2)^2+(y-3)^2=25$
is $S_1+\lambda S_2=0$
$\Rightarrow \quad(1+\lambda) x^2+(1+\lambda) y^2+(6-4 \lambda) x+(4-6 \lambda) y$ $-12(1+\lambda)=0$
$x^2+y^2+\left(\frac{3-2 \lambda}{1+\lambda}\right) x+2\left(\frac{2-3 \lambda}{1+\lambda}\right) y-12=0$ $\ldots(\mathrm{i})$
Cuts the another given circle $(x+1)^2+(y-2)^2=16$ Orthogonally, so
$2\left(\frac{3-2 \lambda}{1+\lambda}\right)-4\left(\frac{2-3 \lambda}{1+\lambda}\right)=-12-11$
$\Rightarrow \quad 6-4 \lambda-8+12 \lambda=-23-23 \lambda$
$\Rightarrow \quad 31 \lambda=-21 \Rightarrow \lambda=-\frac{21}{31}$
$\therefore$ Centre of the circle $(\mathrm{i})$ is
$\left(-\frac{3+\frac{42}{31}}{1-\frac{21}{31}},-\frac{2+\frac{63}{31}}{1-\frac{21}{31}}\right)=\left(-\frac{135}{10},-\frac{125}{10}\right)$
$=\left(-\frac{27}{2},-\frac{25}{2}\right)$
Hence, the correct answer is option: 'A'.