Question

If a circle passing through the point $(-4,3)$ also touches the lines $x+y=2$ and $x-y=2$, then which of the following is/are true?

• A. The centre of the circle is $(-10+3\sqrt{6},0)$
• B. The centre of the circle is $(-10-3\sqrt{6},0)$
• C. The radius of the circle is $r=6\sqrt{2}-3\sqrt{3}$ units
• D. The radius of the circle is $r=6\sqrt{2}+3\sqrt{3}$ units

Answer 1

Answer: (A), (B), (C), (D)

The radius of the circle is $r=6\sqrt{2}+3\sqrt{3}$ units

Given lines are $x+y=2$ and $x-y=2$ which are perpendicular in nature,


We know that,
$r=\left|\frac{h-k-2}{\sqrt{2}}\right|=\left|\frac{h+k-2}{\sqrt{2}}\right|=\sqrt{(h+4{)}^2+(k-3{)}^2}$
Now, using the above conditions
$\left|\frac{h-k-2}{\sqrt{2}}\right|=\left|\frac{h+k-2}{\sqrt{2}}\right|$
By observation, point $(-4,3)$ and centre of the circle lies on the same side of both the lines, so
$-\left(\frac{h-k-2}{\sqrt{2}}\right)=-\left(\frac{h+k-2}{\sqrt{2}}\right)\Rightarrow k=0$
Now,
$\left|\frac{h+k-2}{\sqrt{2}}\right|=\sqrt{(h+4{)}^2+(k-3{)}^2}\Rightarrow \left|\frac{h-2}{\sqrt{2}}\right|=\sqrt{(h+4{)}^2+9}$
Squaring on both sides, we get
$\Rightarrow (h-2{)}^2=2(h^2+8h+25)\Rightarrow h^2+20h+46=0\Rightarrow h=\frac{-20\pm \sqrt{400-184}}{2}\Rightarrow h=-10\pm 3\sqrt{6}$

So, the centre of the circle are
$(-10+3\sqrt{6},0)\text{and}(-10-3\sqrt{6},0)$
Therefore,
$r=\frac{|-10+3\sqrt{6}-2|}{\sqrt{2}}\text{and}r=\frac{|-10-3\sqrt{6}-2|}{\sqrt{2}}\Rightarrow r=|-6\sqrt{2}+3\sqrt{3}|\text{and}r=|-6\sqrt{2}-3\sqrt{3}|\therefore r=6\sqrt{2}-3\sqrt{3}\text{and}r=6\sqrt{2}+3\sqrt{3}$