Question

Find the equation of a circle passing through the points (5, 7), (6, 6) and (2, -2). Find its centre and radius.

Answer 1

Let the required equation of the circle be

$x^2+y^2+2gx+2fy+c=0....\left(i\right)$

Since it passes through each of the points (5, 7), (6, 6) and (2, -2), each one of these points must satisfy (i).

$\therefore 25+49+10g+14f+c=0\Rightarrow 10g+14f+c+74=0...\left(ii\right)$

$36+36+12g+12f+c=0\Rightarrow 12g+12f+c+72=0...\left(iii\right)$

$4+4+4g-4f+c=0\Rightarrow 4g-4f+c+8=0...\left(iv\right)$

Subtracting (ii) from (iii), we get

$2g-2f-2=0\Rightarrow g-f=1....\left(v\right)$

Subtracting (iv) from (iii), we get

$8g+16f+64=0\Rightarrow g+2f=-8....\left(vi\right)$

Solving (v) and (vi), we get $g=-2$ and $f=-3.$

Putting $g=-2$ and $f=-3$ in (ii), we get $c=-12.$

Putting $g=-2,f=-3$ and $c=-12$ in (i), we get

$x^2+y^2-4x-6y-12=0,$

which is the required equation of the circle.

Centre of this circle = $(-g,-f)=(2,3).$

And, its radius = $\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=\sqrt{25}=5$ units.