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Find the equation of a circle passing through the intersection of 4x+y-4=0 and x-y-6=0 with center (-1,-3)

Sagot :

(x+1)^2 +(y+3)^2 =8.08
4x+y-4=0 
x-y-6=0   

Using the process of elimination, we will get x=2 and y=-4. Therefore, the circle passes through point (2, -4).

To find the radius of the circle, use the distance formula.
d = [tex] \sqrt{(2-(-1))² + (-4-(-3))²} [/tex] = √10

The equation of a circle with centre at (h,k) and radius r is 
(x-h)² + (y-k)² = r²

Therefore, the equation of the circle is 
(x+1)² + (y+3)² = 10    

In general form, 

x² +2x +y² +6y = 0