Question

The equation of the circle passing through (1,1)
and through the point of intersection of circles X^2+y^2+ 13x -3y=0 and 2x^2+ 2y^2 +4x -7y-25=0 is:

Answer 1

Answer:

solved

Step-by-step explanation:

solving for  the point of intersection of circles X^2+y^2+ 13x -3y=0 and 2x^2+ 2y^2 +4x -7y-25=0

we have

X²+y²+ 13x -3y=0

 x²+  y² + 2 x -3.5y-12.5=0

11x + 0.5y + 12.5 = 0

22x +  y + 25 = 0

y = - (22x+25)

(22x+25)² + x² + 13x + 66x + 75 = 0

485x² + 1179x + 75 = 0

x² + 2.431 x + 0.155 = 0

x = - 4.1 , 1.7

y = 15.2 , - 62.4

the points of intersection are

(  - 4.1 , 15.2  )   ,  ( 1.7 , - 62.4   ) ≈ (  - 4  , 15  )   ,  ( 2 , - 62  )

adding the 3rd. point  (1,1) we get the required equation.

after a dull algebraic calculations the equation is found to be

x² + y² +2x + y -248 + 243/201(77x+6y+118) = 0