How do you find the equation in general form for the circle centered at (-3,4) and passing through the origin?

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Answer 1 · 2018-05-09T05:18:37

#x^2 + 6x + y^2 -8y = 0#

Given: circle centered at #(-3, 4)# and passing through #(0, 0)#

The standard form of a circle is #(x-h)^2 + (y-k)^2 = r^2#

where #(h, k)# is the center and #r = # radius

Since the circle passes through #(0, 0)#, the radius can be calculated by find the distance between the center and the origin:

#r = sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5#

The circle in standard form is: #(x+3)^2 + (y-4)^2 = 25#

To put the circle in general form you need to distribute.

Use the square rule: #(a + b)^2 = a^2 + 2ab + b^2#

#x^2 + 6x + 9 + y^2 -8y + 16 - 25 = 0#

Add like terms to simplify:

#x^2 + 6x + y^2 -8y = 0#