Circle Ques 82
- A circle touches the line $y=x$ at a point $P$ such that $O P=4 \sqrt{2}$, where $O$ is the origin. The circle contains the point $(-10,2)$ in its interior and the length of its chord on the line $x+y=0$ is $6 \sqrt{2}$. Determine the equation of the circle.
(1990, 5M)
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Answer:
Correct Answer: 82.$x^{2}+y^{2}+18 x-2 y+32=0$
Solution:
Formula:
- The parametric form of $O P$ is
$$ \frac{x-0}{\cos 45^{\circ}}=\frac{y-0}{\sin 45^{\circ}} $$
Since, $\quad O P=4 \sqrt{2}$
So, the coordinates of $P$ are given by
$$ \frac{x-0}{\cos 45^{\circ}}=\frac{y-0}{\sin 45^{\circ}}=-4 \sqrt{2} $$
So, $\quad P(-4,-4)$
Let $C(h, k)$ be the centre of circle and $r$ be its radius.
Now, $C P \perp O P$
$$ \begin{array}{lc} \Rightarrow & \frac{k+4}{h+4} \cdot(1)=-1 \\ \Rightarrow & k+4=-h-4 \\ \Rightarrow & h+k=-8 \\ \text { Also, } & C P^{2}=(h+4)^{2}+(k+4)^{2} \\ \Rightarrow & (h+4)^{2}+(k+4)^{2}=r^{2} \end{array} $$
In $\triangle A C M$, we have
$$ \begin{alignedat} & A C^{2}=(3 \sqrt{2})^{2}+\frac{h+k}{\sqrt{2}} \\ & \Rightarrow \quad r^{2}=18+32 \\ & \Rightarrow \quad r=5 \sqrt{2} \\ & \text { Also, } \quad C P = r \\ & \Rightarrow \quad \frac{h-k}{\sqrt{2}}=r \\ & \Rightarrow \quad h-k= \pm 10 \end{aligned} $$
From Eqs. (i) and (iv), we get
$$ \text { or } \quad(h=1, k=-9) $$
Thus, the equations of the circles are
$$ \begin{array}{rlrl} \text { or } & & (x-1)^{2}+(y+9)^{2} & =(5 \sqrt{2})^{2} \\ \Rightarrow & x^{2}+y^{2}+18 x-2 y+32 & =0 \\ \text { or } & x^{2}+y^{2}-2 x+18 y+32 & =0 \end{array} $$
Clearly, $(-10,2)$ lies interior to
$$ x^{2}+y^{2}+18 x-2 y+32=0 $$
Hence, the required equation of a circle is
$$ x^{2}+y^{2}+18 x-2 y+32=0 $$
BETA
