How do you find the equation of a circle in standard form given x^2+y^2-10x+6y+30=0x2+y2−10x+6y+30=0?

1 Answer
Shwetank Mauria
Jan 23, 2017

(x-5)^2+(y-(-3))^2=2^2(x−5)2+(y−(−3))2=22

Explanation:

The standard form of equation of a circle is the center-radius form of its equation and it is in the format

(x-h)^2+(y-k)^2=r^2(x−h)2+(y−k)2=r2, whose center is (h,k)(h,k) and radius is rr.

To convert x^2+y^2-10x+6y+30=0x2+y2−10x+6y+30=0 into standard form, we should group xx-terms and yy-terms separately as follows:

x^2-10x+y^2+6y=-30x2−10x+y2+6y=−30 and now competing squares, this becomes

(x^2-2xx5xx x+5^2)+(y^2+2xx3xxy+3^2)=-30+5^2+3^2(x2−2×5×x+52)+(y2+2×3×y+32)=−30+52+32

or (x-5)^2+(y+3)^2=-30+25+9(x−5)2+(y+3)2=−30+25+9

i.e. (x-5)^2+(y-(-3))^2=4(x−5)2+(y−(−3))2=4

(x-5)^2+(y-(-3))^2=2^2(x−5)2+(y−(−3))2=22

Hence, center is (5,-3)(5,−3) and radius is 22
graph{x^2+y^2-10x+6y+30=0 [-5, 15, -7.96, 2.04]}

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