Question

Find the equation of the circle passing through the points $(4,1)$ and $(6,5)$ and whose centre is on the line $4x+y=16$

Answer 1

Let the equation of circle be

$x^2+y^2+2gx+2fy+c=0$ .....(1)

where $(-g,-f)$ is center and $g^2+f^2-c$ is the radius.

Since, the circle passes through $(4,1)$, so it satisfies eqn (1)

$\Rightarrow 16+1+8g+2f+c=0$

$\Rightarrow 8g+2f+c=-17$ .....(2)

Since, the circle passes through $(6,5)$, so it satisfies eqn (1)

$\Rightarrow 36+25+12g+10f+c=0$

$\Rightarrow 12g+10f+c=-61$ .....(3)

Subtracting eqn (3) from eqn (2), we get

$-4g-8f=44$

$\Rightarrow g+2f=-11$ ....(4)

Given center lies on the line $4x+y=16$

Since, center $(-g,-f)$ satisfies this equation.

$\Rightarrow -4g-f=16$ ....(5)

Solving eqn (4) and (5), we get

$g=-3,f=-4$

Put this value in (2), we get

$c=15$

Substituting these values in (1),

$x^2+y^2-6x-8y+15=0$

which is the equation of required circle.