How to find the standard form of the equation of the specified circle given that it passes through the origin and has its center at (-5, 4)?

Question

2071 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-05T08:14:29

$x^{2} + y^{2} + 10 x - 8 y = 0$ $x^2+y^2+10x-8y=0$

the eqn of a circle centre $(a, b)$ $(a,b)" "$ and radius $r$ $r$

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

we need to find the radius

we can use Pythagoras

since the circle passes through the origin we have

$r^{2} = {(0 - - 5)}^{2} + {(0 - 4)}^{2}$ $r^2=(0--5)^2+(0-4)^2$

$r^{2} = 25 + 16 = 41$ $r^2=25+16=41$

$:. " eqn "(x- -5)^2+(y-4)^2=41$

$(x+5)^2+(y-4)^2=41$

$x^2+10x+25+y^2-8y+16=41$

$=>x^2+y^2+10x-8y=0$