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Sagot :
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
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
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