Circle

Geometry Level pending

Find the equation of a circle having its center at ( − 3 , 5 ) (-3,5) and a radius of length 4 units.

x 2 + y 2 + 6 x − 10 y + 18 = 0 x^2 + y^2 + 6x - 10y + 18 = 0 x 2 + y 2 − 6 x + 10 y + 18 = 0 x^2 + y^2 - 6x + 10y + 18 = 0 x 2 + y 2 + 10 x − 6 y + 18 = 0 x^2 + y^2 + 10x - 6y + 18 = 0 x 2 + y 2 − 10 x + 6 y + 18 = 0 x^2 + y^2 - 10x + 6y + 18 = 0

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1 solution

The standard form of the equation of a circle is ( x − h ) 2 + ( y − k ) 2 = r 2 (x - h)^2 + (y - k)^2 = r^2 . Substituting − 3 -3 for h h , 5 5 for k k and 4 4 for r r in the standard form and simplfying, we obtain

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

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

x 2 + 6 x + 9 + y 2 − 10 y + 25 = 16 x^2 + 6x + 9 + y^2 - 10y + 25 = 16

x 2 + y 2 + 6 x − 10 y + 18 = 0 x^2 + y^2 + 6x - 10y + 18 = 0

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