How do you write the equation of the circle with Center at (4, 4); passing through (7, 14)?

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Answer 1 · 2016-10-19T10:32:23

Equation of circle is $x^{2} + y^{2} - 8 x - 8 y - 77 = 0$ $x^2+y^2-8x-8y-77=0$

As the circle has center at $(4, 4)$ $(4,4)$ and passes through $(7, 14)$ $(7,14)$ ,

distance between a point $(x, y)$ $(x,y)$ and $(4, 4)$ $(4,4)$ and between $(7, 14)$ $(7,14)$ and $(4, 4)$ $(4,4)$ will be equal. Hence,

$\sqrt{{(x - 4)}^{2} + {(y - 4)}^{2}} = \sqrt{{(7 - 4)}^{2} + {(14 - 4)}^{2}}$ $sqrt((x-4)^2+(y-4)^2)=sqrt((7-4)^2+(14-4)^2)$

or squaring ${(x - 4)}^{2} + {(y - 4)}^{2} = {(7 - 4)}^{2} + {(14 - 4)}^{2}$ $(x-4)^2+(y-4)^2=(7-4)^2+(14-4)^2$

or $x^{2} - 8 x + 16 + y^{2} - 8 y + 16 = 3^{2} + 10^{2}$ $x^2-8x+16+y^2-8y+16=3^2+10^2$

or $x^{2} + y^{2} - 8 x - 8 y + 32 - 9 - 100 = 0$ $x^2+y^2-8x-8y+32-9-100=0$

or $x^{2} + y^{2} - 8 x - 8 y - 77 = 0$ $x^2+y^2-8x-8y-77=0$