Circle 4 Question 22
22. Find the equations of the circles passing through $(-4,3)$ and touching the lines $x+y=2$ and $x-y=2$.
$(1982,3 M)$
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Answer:
Correct Answer: 22. $x^{2}+y^{2}+2(10 \pm 3 \sqrt{6}) x+(55 \pm 24 \sqrt{6})=0$
Solution:
- Let the equation of the required circle be
$ x^{2}+y^{2}+2 g x+2 f y+c=0 $
It passes through $(-4,3)$.
$ \therefore \quad 25-8 g+6 f+c=0 $
Since, circle touches the line $x+y-2=0$ and $x-y-2=0$.
$ \therefore \quad \frac{-g-f-2}{\sqrt{2}}=\frac{-g+f-2}{\sqrt{2}}=\sqrt{g^{2}+f^{2}-c} $
$ \begin{array}{rlrl} \text { Now, } & & \frac{-g-f-2}{\sqrt{2}} & =\frac{-g+f-2}{\sqrt{2}} \\ \Rightarrow & -g-f-2 & = \pm(-g+f-2) \\ \Rightarrow & -g-f-2 & =-g+f-2 \\ \text { or } & -g-f-2 & =g-f+2 \\ \Rightarrow & f & =0 \text { or } g=-2 \end{array} $
Case I When $f=0$
From Eq. (iii), we get
$ \begin{aligned} \frac{-g-2}{\sqrt{2}} & =\sqrt{g^{2}-c} \\ \Rightarrow \quad g^{2}-4 g-4-2 c & =0 \end{aligned} $
On putting $f=0$ in Eq. (ii). we get
$ 25-8 g+c=0 $
Eliminating $c$ between Eqs. (iv) and (v), we get
$ \Rightarrow \quad \begin{aligned} g^{2}-20 g+46 & =0 \\ g & =10 \pm 3 \sqrt{6} \text { and } c=55 \pm 24 \sqrt{6} \end{aligned} $
On substituting the values of $g, f$ and $c$ in Eq. (i), we get
$ x^{2}+y^{2}+2(10 \pm 3 \sqrt{6}) x+(55 \pm 24 \sqrt{6})=0 $
Case II When $g=-2$
From Eq. (iii), we get
$\begin{array}{lc}\Rightarrow & f^{2}=2\left(4+f^{2}-c\right) \ \Rightarrow & f^{2}-2 c+8=0\end{array}$
On putting $g=-2$ in Eq. (ii), we get
$ c=-6 f-41 $
On substituting $c$ in Eq. (vi), we get
$ f^{2}+12 f+90=0 $
This equation gives imaginary values of $f$.
Thus, there is no circle in this case.
Hence, the required equations of the circles are
$ x^{2}+y^{2}+2(10 \pm 3 \sqrt{6}) x+(55 \pm 24 \sqrt{6})=0 $
