Question

If tangents are drawn from a point $P(-3,4)$ to the circle $x^2+y^2=4$, touching at $Q$ and $R$, then the equation of the circumcircle to the triangle $PQR$ is

Answer 1

We have,

Equation of circle is $x^2+y^2=4$

And point $P(-3,4)$

So,

Then,

$x^2+y^2=4$ is a circle with the radius $2$ and is centered at the origin O.

Now, the circumcentre C of $\Delta PQR$ will lie somewhere on the line segment OP. Let C be at distance $c$ from the origin .

Let p be at distance p from the origin. Then, we cen write the equation of the cirumcircle as

Now,

${(x-\frac{(-3)c}{p})}^2+{(y-\frac{4c}{p})}^2={(p-c)}^2$

Where

$p=\sqrt{{(-3)}^2+4^2}$

$p=5$

For simplifying the problem, let’s first rotate the coordinate system about the origin,

So, that P rotated becomes $P^{\prime }=(p,0)=(5,0)$

Which lies on the X-axis.

So,$\Delta PQR$ becomes $\Delta P^{\prime }{Q}^{\prime }{R}^{\prime }$ and it’s circumcentre C becomes $C^{\prime }=(c,0)$.

Let $Q^{\prime }=(2\cos \theta ,2\sin \theta )$

And let $u=\cos \theta$

So,

$u^{\prime }=(2u,2\sqrt{1-u^2})$

Now, using Pythagoras theorem,

$P^{\prime }Q{{}^{\prime }}^2=OP{{}^{\prime }}^2-OQ{{}^{\prime }}^2$

Which means that,

${(5-2u)}^2+{\left(2\sqrt{1-u^2}\right)}^2=5^2-2^2$

So,

$u=\frac{2}{5}$

Now,

$Q^{\prime }=(\frac{4}{5},2\times \frac{\sqrt{21}}{5})$

Since,

$C^{\prime }{P}^{\prime }=C^{\prime }{Q}^{\prime }$

We have,

$(5-c)=\sqrt{{(\frac{4}{5}-c)}^2+{(2\times \frac{\sqrt{21}}{5})}^2}$

$(5-c)=\sqrt{{(\frac{4}{5}-c)}^2+\frac{84}{25}}$

This gives us ,

$c=\frac{5}{2}$

Now, we get the equation of the circumcircle as

${(x+\frac{3}{2})}^2+{(y-\frac{4}{2})}^2=\frac{{(-3)}^2+4^2}{4}$

$\Rightarrow {(x+\frac{3}{2})}^2+{(y-2)}^2=\frac{25}{4}$

Hence, this is the required answer.