How do you find the equation of the circle centered at (h,k) = (-4,-2) and passing through the point (-1,2)?

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Answer 1 · 2016-09-15T06:12:21

$x^{2} + y^{2} + 8 x + 4 x - 5 = 0$ $x^2+y^2+8x+4x-5=0$

The reqd. circle passes thro. the pt. $P (- 1, 2)$ $P(-1,2)$ & has Centre

$C (- 4, - 2)$ $C(-4,-2)$ .

Hence, its radius $r$ $r$ is given by the dist. $C P$ $CP$ , where,

$r^{2} = {(- 1 + 4)}^{2} + {(2 + 2)}^{2} = 9 + 16 = 25$ $r^2=(-1+4)^2+(2+2)^2=9+16=25$ .

Therefore, the eqn. of circle is,

${(x + 4)}^{2} + {(y + 2)}^{2} = 25$ $(x+4)^2+(y+2)^2=25$

$:. x^2+y^2+8x+4x+16+4-25=0,$ " i.e.,

$x^2+y^2+8x+4x-5=0$