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

equation of circle $={(x-h)}^2+{(y-k)}^2=r^2$

Point $(4,1)$ and $(6,5)$ passing from circle

so point satisfy the condition

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

$16-8h+h^2+1-2k+k^2=r^2$

$h^2+k^2-8h-2k+17=r^2.......\left(i\right)$

point $(6,5)$

$\Rightarrow$ ${(6-h)}^2+{(5-k)}^2=r^2$

$36-12h+h^2+25-10k+k^2=r^2$

$\therefore$ $h^2+k^2-12h-10k+61=r^2.......\left(ii\right)$

Compare (i) and (ii)

$h^2+k^2-8h-2k+17=h^2+k^2-12k-10k+61$

$\therefore$ $12h-8h+10k-2k+17-61=0$

$4h+8k-4h=0$

$4h+8k=4h$

$h+2k=\mathrm{11.......}\left(iii\right)$

given center of the circle on line $4x+y=16$

$\therefore$ $4h+k=\mathrm{16....}\left(iv\right)$

solve equation (iii) and (iv)

$4h+8k=44$

$4h+k=16$

-----------------------

$7k=28$

$k=4$

$h=11-8=3$

Substitute value of $h$ and $k$ in (i)

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

$r^2=7+9=10$

$r=\sqrt{10}$

equation of circle

${(x-h)}^2+{(y-k)}^2=r^2$

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