Question

A circle passes through the points $(-1,1),(0,6)$, and $(5,5)$. Find the points on this circle the tangents at which are parallel to the straight line joining the origin to its centre.

Answer 1

Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$

It passes through $(-1,1)$

$\Rightarrow 1+1-2g+2f+c=0\Rightarrow -2g+2f+c=-\mathrm{2......}\left(i\right)$

It passes through $(0,6)$

$\Rightarrow 0+36+0+12f+c=0\Rightarrow 12f+c=-\mathrm{36......}\left(ii\right)$

It passes through $(5,5)$

$\Rightarrow 25+25+10g+10f+c=0\Rightarrow 10g+10f+c=-\mathrm{50.......}\left(iii\right)$

Subtracting $\left(ii\right)$ from $\left(i\right)$ we get

$-2g-10f=\mathrm{34.......}\left(iv\right)$

Subtracting $\left(iii\right)$ from $\left(ii\right)$ we get

$-10g+2f=\mathrm{14.......}\left(v\right)$

Solving $\left(iv\right)$ and $\left(v\right)$

$\Rightarrow g=-2,f=-3$

Substituting $g$ and $f$ in $\left(i\right)$ we get $c=0$

So the equation of circle is

$x^2+y^2-4x-6y=\mathrm{0.................}\left(vi\right)$

Centre of the circle is $(2,3)$

Slope of line joining centre to origin $=\frac{3-0}{2-0}=\frac{3}{2}$

Equation of line passing through the centre and perpendicular to line joining centre to origin is

$y-3=-\frac{2}{3}(x-2)3y-9=-2x+42x+3y=\mathrm{13........}\left(vii\right)$

Tangents will be parallel to the line joining the origin to the centre at the points where $\left(vii\right)$ cuts the circle

Substituting $y$ from $\left(vii\right)$ in $\left(vi\right)$ we get

$13x^2-52x-65=0x^2-4x-5=0x^2-5x+x-5=0x(x-5)+(x-5)=0(x+1)(x-5)=0x=-1,5$

Substituting $x$ in $\left(vii\right)$ we get

$y=5,1$

So the point of contact of tangents are $(-1,5)$ and $(5,1)$