# How do you find the equation of a circle passing through the points of intersection of the line x+y=4 and the circle x^2+y^2=16 and the point (10,2)?

Sep 21, 2016

${x}^{2} + {y}^{2} - 11 x - 11 y + 28 = 0$.

#### Explanation:

Method I :-

I will outline the soln.

First, solving the eqns. of the line and the circle, we will get the co-ords. of two pts. of intersection. Using these two pts. together with the given pt. $\left(10 , 2\right)$, we can find the eqn. of reqd. circle.

Method II :-

This Method is simpler than the Method I.

In this Method, we use the following Result :

Result :- If a circle $S : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$ and a line

$L : a x + b y + c = 0$ intersect, then the eqn. $S + k L = 0 , k \in \mathbb{R}$

represents a circle passing through their pts. of intersection.

So, let us suppose that the reqd. eqn. of circle is,

${x}^{2} + {y}^{2} - 16 + k \left(x + y - 4\right) = 0 , k \in \mathbb{R}$.

This circle passes thro. $\left(10 , 2\right)$. therefore,

${10}^{2} + {2}^{2} - 16 + k \left(10 + 2 - 4\right) = 0 \Rightarrow 88 + k \left(8\right) = 0.$

$\Rightarrow k = - 11$. Hence, the reqd. eqn. is,

${x}^{2} + {y}^{2} - 16 - 11 x - 11 y + 44 = 0$, or,

${x}^{2} + {y}^{2} - 11 x - 11 y + 28 = 0$.

Enjoy Maths.!