Circle 4 Question 11
11. The equation of the circle passing through $(1,1)$ and the points of intersection of $x^{2}+y^{2}+13 x-3 y=0$ and $2 x^{2}+2 y^{2}+4 x-7 y-25=0$ is
(1983, 1M)
(a) $4 x^{2}+4 y^{2}-30 x-10 y=25$
(b) $4 x^{2}+4 y^{2}+30 x-13 y-25=0$
(c) $4 x^{2}+4 y^{2}-17 x-10 y+25=0$
(d) None of the above
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Answer:
Correct Answer: 11. (b)
Solution:
- The required equation of circle is
$\left(x^{2}+y^{2}+13 x-3 y\right)+\lambda \quad 11 x+\frac{1}{2} y+\frac{25}{2}=0$
Its passing through $(1,1)$,
$$ \begin{array}{rlrl} \Rightarrow & 12+\lambda(24) & =0 \\ \Rightarrow & & \lambda & =-\frac{1}{2} \end{array} $$
On putting in Eq. (i), we get
$$ \begin{aligned} & & x^{2}+y^{2}+13 x-3 y-\frac{11}{2} x-\frac{1}{4} y-\frac{25}{4} & =0 \\ \Rightarrow & & 4 x^{2}+4 y^{2}+52 x-12 y-22 x-y-25 & =0 \\ \Rightarrow & & 4 x^{2}+4 y^{2}+30 x-13 y-25 & =0 \end{aligned} $$
