# What is the center and the radius of the circle x^2 + y^2 -3x + 5y -25 = 0

We can write the equation which we have been given: x^2 + y^2 -3x + 5y -25 = 0 as the general equation of a circle.

x^2 + y^2 -3x + 5y -25 = 0

=> x^2 -3x + y^2 + 5y -25 = 0

=> x^2 - 3x + (3/2)^2 + y^2 + 5y + (5/2)^2 = 25 + (3/2)^2 + (5/2)^2

=> (x - 3/2)^2 + (y + 5/2)^2 = 33.5

Therefore the center of the circle is (3/2, -5/2) and radius is sqrt 33.5.

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x^2 + y^2 -3x + 5y -25 = 0

We are given the equation of the circle.

We need to determine the radius and the center of the circle.

First, we need to rewrite the equation into the standard form:

(x-a)^2+ (y-b)^2 = r^2  such that (a,b) is the center and r is the radius.

Then we will complete the squares.

==> x^2 -3x + y^2 +5y = 25

=> x^2 -3x + 9/4 + y^2 + 5y + 25/4 = 25 + 9/4 + 25/4

==> (x -3/2)^2 + (y+5/2)^2 = (100+9+25)/4

==> (x-3/2)^2 + (y+5/2)62 = 134/4 = 33.5

==> (x-3/2)^2 + (y+5/2)^2 = 33.5

Then, the center of the circle is the point (3/2, -5/2)

and the radius is sqrt(134) /2 = sqrt(33.5) = 5.79 (approx).

Approved by eNotes Editorial Team