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

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### 3 Answers

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.**

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).**

The equation of a circle with centre (h,k) and radius r is given by: (x-h)^2+(y-k)^2 = r^2, where (x,y) is any point on the circumference of the circle.

So we convert the given equation x^2 + y^2 -3x + 5y -25 = 0 of the circle in the form (x-h)^2+(y-k)^2= r^2:

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

x^2-3x+y^2+5y = 25. We add (3/2)^2+(5/2)^2 to both sides .

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 = 25+(9+25)/4 = 134/4 = {(sqrt134)/2}^2.

(x-3/2)^2+(y-(-5/2))^2 = {(sqrt134)/2}^2.....(2). Now if we identify this with the circle (x-h)^2+(y-k)^2 = r^2, we get:

(h,k) = (3/2, -5/2) and r = (1/2)sqrt(134).

Therefore the centre of the given circle is (3/2, -5/2) and the radius = (1/2) sqrt(134).