Find an equation of the circle centered at (6,-5) and passing through (1,7).?

Answer 1

First figure the distance between the two points:

d = sqrt(Δx² + Δy²)
Δx = x1 - x2 = 6 - 1 = 5
Δy = y1 - y2 = -5 - 7 = -12
d = sqrt(5² + (-12)²)
d = sqrt(25 + 144)
d = sqrt(169)
d = 13

When you have a point (x1, y1) as the center and r as a radius, the general formula for a circle is:

(x - x1)² + (y - y1)² = r²

So using your values of:
x1 = 6
x2 = -5
r = 13

(x - 6)² + (y - (-5))² = 13²

This simplifies to:
(x - 6)² + (y + 5)² = 169

Answer 2

First, the equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where the center of the circle is the point (h,k) and r is the radius.
That gives us (x-6)^2 + (y+5)^2 = r^2.
So, we have to determine the radius. We know the center point, and we have one point in the circle. If we graph those point on graph paper, we can construct a right triangle with the line connecting those 2 points as our hypotenuse. The base of that triangle would be 5 and the height would be 12. Using the rules we know about right triangles, also known as the Pythagorean Theorem, we've got 5^2 + 12^2 = c^2, c being our hypotenuse. 25 + 144 = c^2, or 169 = c^2, giving us c = 13. That's our radius.

So, the equation for this circle is (x-6)^2 + (y =5)^2 = 169

Answer 3

The general equation of a circle is

(x-xc)^2 + (y-yc)^2 = r^2

where (xc, yc) is the center of the circle and r is the radius.

So you already know the center. The radius can be found by plugging the point on the circle into the equation and solving for r

Answer 4

(x - 6)^2 + (y + 5)^2 = 169

Answer 5

(x-6)^2 + (y+5)^2 = 169

Answer 6

(x-x1)^2+(y-y1)^2=r^2
(x-6)^2+(y-(-5))^2=r^2
(x-6)^2+(y+5)^2=r^2
r^2=(y2-y1)^2+(x2-x1)
r^2=(7+5)^2+(1-6)^2
r^2=12^2+5^2
r^2=144+25=169

Here's the answer:
(x-6)^2+(y+5)^2=169