Question

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

Answer 1

Let the equation of required circle be $(x-b{)}^2+(y-k{)}^2=r^2$.. (1)

It gives that circle passes through $A(4,1)$ & $B(6,5)$

$(4,h{)}^2+(1-k{)}^2=r^2\&(6-h{)}^2+(5-k{)}^2=r^2$

$(4-h{)}^2+(1-k{)}^2=(6-h{)}^2+(5-k{)}^2$

$16-8h+h^2+1-2k+k^2=36-12h+h^2+25-10k+k^2$

$16-8h+1-2k=36-12h+25-10k$

$4h+8k=44$

$h+2k=11$ ... (2)

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

$\therefore 4h+k=16$ ... (3)

Solving (2) & (3) $-7k=-28$

$k=4$

i.e; $4h+4=16\to h=3$

substitute h, k in (1)

$(4-3{)}^2+(1+4{)}^2=r^2$

$1^2+(-3{)}^2=r^2$

$r^2=10$ or $r=\sqrt{10}$

Hence the equation of circle is

$(x-3{)}^3+(y-4{)}^2=(\sqrt{10}{)}^2$

$x^2-6x+9+y^2-8y+16=10$

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

is the required equation of circle.