Question

Find the centres of the circles passing through $(-4,3)$ and touching the lines $x+y=2$ and $x-y=2$.

• A. $\left(\right(-10\pm \sqrt{54}),0)$
• B. $(10\pm \sqrt{54},0)$
• C. $(0,-10\pm \sqrt{54})$
• D. $(0,10\pm \sqrt{54})$

Answer 1

Answer: (A)

$\left(\right(-10\pm \sqrt{54}),0)$

Let $AB$ and $AC$ be the straight lines $x+y=2$ and $x-y=2$, respectively
Let $P(h,0)$ be the centre of the circle touching $AB$ and $AC$.
Then, the radius of the circle is given by $r=\sqrt{{(h+4)}^2+9}$
Since this is the distance between the centre $P$ and the point $(-4,3)$
Hence, the equation of the circle is obtained as ${(x-h)}^2+y^2=r^2={(h+4)}^2+9$
Further, the perpendicular distance from $P$ upon the line $AB$ is given by $\frac{h-2}{\sqrt{2}}$
Since this is equal to the radius, we must have $\frac{{(h-2)}^2}{2}={(h+4)}^2+9$
Simplifying, we obtain $h=-10\pm 3\sqrt{6}$
Hence, the equation of the circle with centre $(h,0)$ and radius $r$ is given by
${(x-h)}^2+y^2={(h+4)}^2+9,h=-10\pm 3\sqrt{6}$