A circle and a straight line intersect at (-2 ; 6) and (6 ; -2). Determine the equation of the circle and of the straight line.

Question

(-2;6) and (6;-2)

Answer 1

If the line intersects the circle then it must be a chord of the circle. If we label the centre of the circle O and the ends of the chord A(-2,6) and B(6,-2), we can use simple geometry to find the centre of the circle. AOB is an isosceles triangle where AO=BO=radius of the circle. AO=BO=6-(-2)=8. The centre of the circle is at (6-8,6-8)=(-2,-2).

The equation of the circle is therefore (x-(-2))²+(y-(-2))²=64=(x+2)²+(y+2)².

The slope of AB is (-2-6)/(6-(-2))=-8/8=-1. Therefore the equation of the line is:

y-6=-(x-(-2)), y=6-x-2=4-x.