Equation of the circle that is concentric with x²+y²+4x-6y+4=0 and passes through P(2,6).?

Question

Please help me find an equation of this problem..this is for our assignment in algebra and trigonometry..thanks

Answer 1

First find a proper equation for the circle:

(x^2 + 4x) + (y^2 - 6y) + 4 = 0

(x^2 + 4x + 4) - 4 + (y^2 - 6y + 9) - 9 + 4 = 0

(x + 2)^2 + (y - 3)^2 = 9

So the circle is centered at (-2,3) and has a radius of 3.

You now want a circle concentric to this one that passes through the point (2,6). This will have the same center but a different radius so:

(x + 2)^2 + (y - 3)^ = r^2

(2 + 2)^2 + (6 - 3)^2 = r^2

16 + 9 = 25 = r^2 and r = 5

So the concentric circle will have a radius of 5. The equation is:

(x + 2)^2 + (y - 3)^2 = 25

Answer 2

The terms

x²+y²+4x-6y

are the ones which determine the centre, so a circle with the same centre must have equation

x²+y²+4x-6y + constant = 0

To find the constant, use the fact that the circle passes through (2, 6), so substitute

x = 2, y = 6 in the equation:

2²+6²+4*2-6*6 + constant = 0

12 + constant = 0

constant = -12

Therefore the required equation is

x²+y²+4x-6y - 12 = 0

It may be that you are required to do something fuller than this, so below is a longer solution (but not as wordy!):

x²+y²+4x-6y = -4

Complete the squares:

x² + 4x + 4 + y² - 6y + 9 = -4 + 4 + 9

(x + 2)² + (y - 3)² = 9

Given circle has centre (-2, 3), radius 3.

A circle  concentric with it, with radius r , has equation

(x + 2)² + (y - 3)² = r²

It passes through (2, 6) if

(2 + 2)² + (6 - 3)² = r²

25 = r²

Hence the equation is

(x + 2)² + (y - 3)² = 25

Expanding and collecting terms on the left gives

x²+y²+4x-6y - 12 = 0