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Q.
Find the equation of the circle which passes through the points $(20, 3)$, $(19, 8)$ and $(2, -9)$. Find its centre and radius.
Conic Sections
Solution:
By substitution of coordinates in the general equation of the circle given by
$x^{2}+y^{2} +2gx+2fy+c=0$, we have
$40g + 6f+ c = - 409\, \ldots\left(i\right)$
$38g+ 16f+ c = -425\, \ldots\left(ii\right)$
$4g - 18f+ c = - 85\, \ldots\left(iii\right)$
Solving $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$, we get
$g = - 7 , f= - 3$ and $c = - 111$
Hence, the equation of the circle is
$x^{2}+y^{2} - 14x - 6y - 111 = 0$
or $\left(x - 7\right)^{2} + \left(y - 3\right)^{2} = 13^{2}$
Therefore, the centre of the circle is $\left(7,3\right)$ and radius is $13$.