To find the equation of a circle passing through the point (-5, -2) and with the same center as the circle x^2 + y^2 - 4x + 8y - 40 = 0, we first need to find the center of the given circle.
The general equation of a circle is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.
Given circle equation: x^2 + y^2 - 4x + 8y - 40 = 0
Rearranging terms, we get: (x^2 - 4x) + (y^2 + 8y) = 40
Completing the square for x and y terms:
(x^2 - 4x + 4) + (y^2 + 8y + 16) = 40 + 4 + 16
(x - 2)^2 + (y + 4)^2 = 60
Comparing with the general equation, we find that the center of the given circle is at the point (2, -4).
Since the new circle has the same center, we can use the center as (2, -4) and the point (-5, -2) to find the radius of the new circle:
r = √[(-5 - 2)^2 + (-2 + 4)^2]
r = √(49 + 4) = √53
Now we have the center (2, -4) and the radius √53 of the new circle. We can write the equation of this new circle as:
(x-2)^2 + (y+4)^2 = 53.
Find the equ. of a circle passing through the (- (5, -2) and has the same centre as the circle x ^ 2 + y ^ 2 - 4x + 8y - 40 = 0
1 answer
1 answer