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The equation of the circle passing through (1,1) & points of intersection of x2+y2+13x-3y = 0 & 2x2+2y2 + 4x-7y-25 = 0 is??

The equation of the circle passing through (1,1) & points of intersection of x2+y2+13x-3y = 0 & 2x2+2y2 + 4x-7y-25 = 0 is??

Grade:11

3 Answers

APURV GOEL
39 Points
9 years ago

Take the equation of the circle as x2+y2+13x-3y + k(2x2+2y2 + 4x-7y-25) = 0 where k is an arbitrary real no.

Put the values of the point (1,1) in the equation and find the value of k then rewrite this equation with the value of k and solve to get the eq.

k=1/2

and eqn is 4x2 + 4y2 + 30x - 33y - 25 = 0

Ranjita yadav
35 Points
9 years ago

why u put 1,1 inplace of x & y??

APURV GOEL
39 Points
9 years ago

It is given in the question that the circle we are finding passes through (1,1) so i put them so as to satisfy the equation and find the value of k. Once the value of k is known question is over buddy.

If u liked my answer please approve it.

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