find-the-equation-of-the-circle-containing-the-point-2-2-and-passing-throught-the-points-of-intersection-of-the-two-circle-x-2-y-2-3x-2y-4-0-and-x-2-y-2-2x-y-6-0-

Question Number 95639 by i jagooll last updated on 26/May/20

find the equation of the circle

containing the point (- 2, 2) and

passing throught the points of

intersection of the two circle

x^{2} + y^{2} + 3 x - 2 y - 4 = 0 and

x^{2} + y^{2} - 2 x - y - 6 = 0

Commented by i jagooll last updated on 26/May/20

thank you

Answered by john santu last updated on 26/May/20

substitute (- 2, 2) for (x, y) in the

equation (x^{2} + y^{2} + 3 x - 2 y - 4) + λ (x^{2} + y^{2} - 2 x - y - 6) = 0

then λ = \frac{3}{2} . so desired equation can be

written as {5 x}^{2} + {5 y}^{2} - 7 y - 26 = 0