Question

Find the equation of the circle which passes through the point $(2,-2)$ and $(3,4)$ and whose centre lies on the line $x+y=2$.

Answer 1

We know that the equation of a circle with center at $(h,k)$ is,

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

Now, this circle passes through $(2,-2)$. Then

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

$h^2+k^2-4h+4k+8=r^2$

Again, this circle passes through $(3,4)$. Then

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

$h^2+k^2-6h-8k+25=r^2$

Subtracting the above results, we have

$2h+12k=17$ …… (1)

Now, the center lies on $x+y=2$. So,

$h+k=2$ ….. (2)

Solving equations (1) and (2), we get

$h=\frac{7}{10}$ and $k=\frac{13}{10}$

Put values of $(h,k)$ in the first expression.

${(2-\frac{7}{10})}^2+{(-2-\frac{13}{10})}^2=r^2$

$r^2=\frac{169+1089}{100}$

$r=\frac{\sqrt{1258}}{10}$

Therefore, equation of the circle is,

${(x-\frac{7}{10})}^2+{(y-\frac{13}{10})}^2=\frac{1258}{100}$

Hence, this is the required equation.