How do you find the equation of the circle with center at (3, 2) and through the point (5, 4)?

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Answer 1 · 2016-06-18T21:31:04

The equation is $x^{2} + y^{2} - 6 x - 4 y + 5 = 0$ $x^2+y^2-6x-4y+5=0$ .

The generic equation of a circle with center $(c_{x}, c_{y})$ $(c_x, c_y)$ and radius $r$ $r$ is:

${(x - c_{x})}_{x}^{2} + (y - c_{y}) = r^{2}$ $(x-c_x)^2+(y-c_y)=r^2$

in our case

${(x - 3)}^{2} + {(y - 2)}^{2} = r^{2}$ $(x-3)^2+(y-2)^2=r^2$

What is missing here is the radius, but we can find it substituting the point $(5, 4)$ $(5,4)$

${(5 - 3)}^{2} + {(4 - 2)}^{2} = r^{2}$ $(5-3)^2+(4-2)^2=r^2$

$2^{2} + 2^{2} = r^{2}$ $2^2+2^2=r^2$

$4 + 4 = r^{2}$ $4+4=r^2$

$r = \sqrt{8}$ $r=sqrt(8)$

The final equation is

${(x - 3)}^{2} + {(y - 2)}^{2} = 8$ $(x-3)^2+(y-2)^2=8$ or, expanding the squares,

$x^{2} + y^{2} - 6 x - 4 y + 5 = 0$ $x^2+y^2-6x-4y+5=0$ .