Question

A circle whose center is at (3,-5) has equation

Answer 1

Answer: a = 10

radius = 7

Step-by-step explanation:

The generic equation for a circle of radius R is, centered in the point (a, b)

(x - a)^2 + (x - b)^2 = R^2

In this case, we know that the center is (3, -5)

Then the equation of the circle must be something like:

(x - 3)^2 + (y + 5)^2 = R^2

And we have:

x^2 + y^2 - 6*x + a*y = 15

First, let's complete squares for x:

(x - 3)^2 = x^2  + 2*(-3)*x + (-3)^2 = x^2 - 6*x + 9.

Then we can add 9 to both sides of the equation to get:

x^2 + y^2 - 6*x + a*y + 9 = 15 + 9

(x^2 - 6*x + 9) + y^2 + a*y = 24

(x - 3)^2 + y^2 + a*y = 24

Now let's do the same for y:

(y + 5)^2 = y^2 + 2*5*y + 5^2 = y^2 + 10*y + 25

From this we can already see that a must be equal to 10, if we replace a by 10, the circle equation becomes:

(x - 3)^2 + y^2 + 10*y = 24

Now we can add 25 to both sides of the equation to get:

(x - 3)^2 + y^2 + 10*y + 25 = 24 + 25

(x - 3)^2 + (y + 5)^2 = 49

An the radius of the circle will be equal to:

r = √49 = 7