Question

Find the equation to the circle passing through the points(12, 43), (18, 39), and (42, 3) and prove that it also passes through the points ( - 54, - 69) and ( - 81, - 38).

Answer 1

Let the circle be

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

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

$(12-h{)}^2+(43-k{)}^2=r^2$

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

$h^2+k^2-24h-86k+1993=r^2$ ........(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)

Subtracting (3) from (1) and (2) from (1)

$60h-80k+220=0$

$3h-4k+11=0$ .............(4)

$12h-8k+148=0$

$3h-2k+37=0$ ....................(5)

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

$k=-13,h=-21$

Putting values in (1)

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

So, equation of circle is

$(x+21{)}^2+(y+13{)}^2=(65{)}^2$

For proving that the point lies on the circle, the point must satisfy the equation of circle

For point $(-54,-69)$

$\Rightarrow (-54+21{)}^2+(-69+13{)}^2=(65{)}^2$

$\Rightarrow 4225=4225$

Hence proved

For point $(-81,-38)$

$\Rightarrow (-81+21{)}^2+(-38+13{)}^2=(65{)}^2$

$\Rightarrow 4225=4225$

Hence proved.