# How do you write the equation for a circle through (-2,4), (6,0) and (1,5)?

Aug 25, 2016

We assume the following result :

Result : If a circle $S ' : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0 , \text{and, a line} L :$

$l x + m y + n = 0 \text{ intersect, then} S ' + \lambda L = 0 , \lambda \in \mathbb{R} ,$

represents a circle that passes through their points of intersection.

Consider a circle $S '$ having diametrically opposite pts.

$\left(- 2 , 4\right) \mathmr{and} \left(1 , 5\right)$. Then,

$S ' : \left(x + 2\right) \left(x - 1\right) + \left(y - 4\right) \left(y - 5\right) = 0 ,$, or,

$S ' : {x}^{2} + {y}^{2} + x - 9 y + 18 = 0$

The Eqn. of the line $L$ through these pts.
$L : \det | \left(x , y , 1\right) , \left(- 2 , 4 , 1\right) , \left(1 , 5 , 1\right) | = 0 , i . e . , L : - x + 3 y - 14 = 0$.

We observe that the Reqd. Circle $S$ passes through the pts. of intersection of circle $S '$ and line $L$. Hence, by the above Result,

$S : S ' + \lambda L = 0 :$, i.e.,

$S : {x}^{2} + {y}^{2} + x - 9 y + 18 + \lambda \left(- x + 3 y - 14\right) = 0 , \lambda \in \mathbb{R}$

The pt. $\left(6 , 0\right) \in S$,

$\Rightarrow 36 + 0 + 6 - 0 + 18 + \lambda \left(- 6 + 0 - 14\right) = 0$.

$\Rightarrow 60 - 20 \lambda = 0 \Rightarrow \lambda = 3$. Hence,

$S : {x}^{2} + {y}^{2} - 2 x - 24 = 0$.

Enjoy Maths.!