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)?

1 Answer
Sep 21, 2016

#x^2+y^2-11x-11y+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. #(10,2)#, 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+2gx+2fy+c=0# and a line

#L : ax+by+c=0# intersect, then the eqn. #S+kL=0, k in RR#

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(x+y-4)=0, k in RR#.

This circle passes thro. #(10,2)#. therefore,

#10^2+2^2-16+k(10+2-4)=0 rArr 88+k(8)=0.#

# rArr k=-11#. Hence, the reqd. eqn. is,

#x^2+y^2-16-11x-11y+44=0#, or,

#x^2+y^2-11x-11y+28=0#.

Enjoy Maths.!