Find the equation to the circle passing through the points$(12,43),(18,39)$ and $\left(42,3\right)$

Concept: If a circle passes through the three points, then they must satisfy the equation of the circle.

Let the circle be ${(x-h)}^2+{(y-k)}^2=r^2$

Here,  $(12,43),(18,39),(42,3)$satisfy the equation.

For first point: ${(12-h)}^2+{(43-k)}^2=r^2$

$h^2+144-24h+1849+k^2-86k=r^2$

$h^2+k^2-24h-86k+1993=r^2$  ……(1)

For second point:

${(18-h)}^2+{(39-k)}^2=r^2$

$h^2+k^2-36h-28k+1845=r^2$……(2)

For third point:

${(42-h)}^2+{(3-k)}^2=r^2$

$h^2+k^2-84h-6k+1773=r^2$ ……(3)

When subtracting (3) from (1) and (2) from (1), we get

$60 \text{h}-80\text{k}+220=0$

$3 \text{h}-4\text{k}+11=0$…..(4)

$12 \text{h}-8\text{k}+148=0$

$3 \text{h}-2\text{k}+37=0$…..(5)

When solving (4) and (5), we get

$\text{k}=-13, \text{h}=-21$

When putting values in (1), we get

$r^2={(12+21)}^2+{(13+43)}^2={\left(65\right)}^2$

Hence, equation of the circle is ${(x+21)}^2+{(y+13)}^2={\left(65\right)}^2$