Question

Find the equation of a circle which passes through $(4,1),(6,5)$ and having the centre on $4x+y=16$

Answer 1

The general equation of the circle is $x^2+y^2+2gx+2fy+c=0$.

Given that $(4,1)$ lies on the circle.

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

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

$(6,5)$ lies on the circle.

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

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

Subtracting equations (i) and (ii), we get

$4g+8f=-44$

$g+4f=-11$ ....(iii)

Let $(-g,-f)$ be the centre of the circle lying on $4x+y=16$.

$\therefore -4g-f=16$ .....(iv)

Now, $2$ (iv) $+$ (iii)

$-7g=21$

$\Rightarrow g=-3$

Substituting this value in equation (iv), we get

$-4(-3)-f=16$

$\Rightarrow f=-4$

Substitute in (iv), we get

$8(-3)+2(-4)+c=-17$

$\Rightarrow -24-8+c=-17$

$\Rightarrow c=15$

Therefore, the general equation of the circle is $x^2+y^2-6x-8y+15=0$.