Equation of a Circle which share the same center

Question

How to find the equation of the circle which passes through the point $(-2,-4)$ and has the same center as the circle whose equation is $x^2+ y^2 -4x - 6y -23$ ?

Answer 1

Complete the squares of $$x^2 + y^2 - 4x - 6y - 23 = 0$$ to find it's center. (See the formula for the equation of a circle with radius r and center $(a, b)$, below.) Once you complete the squares for the given circle and obtain that form, you can "read off" the center of both circles.

$$x^2 - 4x + 4 + y^2 - 6y + 9 = 23 + 4 + 9 \iff (x-2)^2 + (y-3)^2 = 36 = 6^2$$

So the center of the circle is $(2, 3)$.

Then use your given point, along with the center you find, to compute the distance from the center to $(-2, -4)$, which will give you the desired circle's radius. If $(a, b)$ is the center of your circle, then $$r = \sqrt{(a - (-2))^2 + (b - (-4))^2}$$

That means the radius of our desired circle is given by $$r = \sqrt{(2 - (-2))^2 + (3 - (-4))^2} = \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$$

Then use the standard form for an equation of a circle to construct the equation of your desired circle: $r:$ radius, $(a, b):$ center gives us $$(x - a)^2 + (y - b)^2 = r^2$$

$$(x - 2)^2 + (y - 3)^2 = 65$$

Answer 2

the equation of a circle is $(x-a)^2+(y-b)^2=r^2$ so get your equation to look like $(x-a)^2+(y-b)^2=23$ then the center of the circle is (a,b) and the equation of the circle you want is $(x-a)^2+(y-b)^2=R^2$ where R is the distance from the center to (-2,-4)

Answer 3

Simply leave unchanged everything except the constant term. Substitute $x=-2$, $y=-4$ to find it.

Answer 4

Take your existing circle and find its center C, by differentiating the equation by $x$ and by $y$

$$ \frac{{\rm d}}{{\rm d}x}(x^2+ y^2 -4x - 6y -23) = 0 \\\frac{{\rm d}}{{\rm d}y}(x^2+ y^2 -4x - 6y -23) = 0 $$

$$ 2x_c-4=0 \\ 2y_c-6=0 $$

$$ x_c = 2 \\ y_c = 3 $$

Find the radius from the distance between C and your point A = (-2,-4)

$$ r = \sqrt{ ((-2)-2)^2 + (-4-3)^2 } = \sqrt{65} $$

Your equation is

$$ \boxed{ (x-x_c)^2 + (y-y_c)^2 - r^2 = 0 }$$