Question 858395
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Hi there,

Problem:
What is the equation of a circle in general form whose center is at (-3,-4) and passes through the point (1,2)?

Solution:
Begin with the equation for a circle in standard form where (a,b) is the center of the circle and r is its radius. 
{{{(x-a)^2 + (y-b)^2=r^2}}}

We are given the center of the circle at (a,b)=(-3,-4), so our equation becomes
{{{(x-(-3))^2 + (y-(-4))^2 = r^2}}}
{{{(x+3)^2 + (y+4)^2 = r^2}}}

We all have a point (1,2) on the circle. Substitute 1 for x and 2 for y into the circle to find the radius r^2.
{{{(1+3)^2 + (2+4)^2 = r^2}}}
{{{16+36 = r^2}}

Substitute 36 for r^2 in the standard form of the equation.
{{{(x+3)^2+(y+4)^2=52}}}

This is the equation for our circle in standard form. The equation for a circle in general form looks like
{{{x^2+y^2+Ax+By+C+0}}}

To translate our equation to general form. Multiply everything out.
{{{x^2+6x+9+y^2+8y+16=52}}}

Subtract 36 from both sides and combine like terms. Then rearrange in the proper order.
{{{x^2+6x+y^2+8y-27=0}}}

The equation in general form is
{{{x^2+y^2+6x+8y-27=0}}}
 
Feel free to email me if your have questions about the solution.

~Mrs. Figgy
math.in.the.vortex@gmail.com
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