# Find the center and radius of each circle. x^2+y^2+4x+6y+9=0 x^2+y^2+10x-14y-7=0 3x^2+3y^2-12x-24y+12=0NO

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The equation of the circle with centre (h,k) and radius r is given by : (x-h)^2 +(y-k)^2 = r^2 . Or

x^2+y^2 -2hx-2ky +(h^2+k^2-r^2 ) = 0.........(1) Now compare this with the equation of the circle:

x^2+y^2+2fx+2gy+C= 9................................(2), Comparing coeffcients in (1) and (2),

(h,k) = (-f, -g) and r^2 = f^2+g^2-c

1) Therfore,

x^2+y^2+4x+6y+9=0 is a circle with centre (-f,-g ) = (-4/2, -6/2) = (-2,-3) .And radius : r^2 = f^2+g^2-c = 2^2+3^2-9 = 4. Therefore, radius r = 2.

2) x^2+y^2+10x-14y-7=0

centre: (-f,-g) = (-10/2 , -(-14/2) ) = (-5, 7)

Radius: r^2 = (-5)^2+7^2- (-7) = 25+49+7 = 81. Or radius r = 81^(1/2) = 9

3)

3x^2+3y^2-12x-24y+12=0 .We Bring it to standard rd form by making the coefficients of x^2 and y^2 to unity. So we divide the equation by 3 and rewrite as:

x^2+y^2 -(12/3)x - (24/3)y +12/3 = 0 Or

x^2+y^2-4x-8x+4 = 0.

Cenre: (-f,-g) = (--4/2, --8/2) = (2,4)

Radius: r^2 = f^2+g^2 -c = 2^2+4^2-4 =16. So radius, r = 4.