Question

Write the standard form of the equation of the circle center (0,4) passes through the point (-2,-4)

Answer 1

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As we know, the standard form of circle is written in this way ~

$\qquad \sf x{}^{2} + y {}^{2} + 2gx + 2fy + c = 0$

where, the coordinates of centre is (-g , -f) and radius equals to :

$\qquad \sf \dashrightarrow \: \sqrt{g {}^{2} + {f}^{2} - c }$

Now, it's time to equate the coordinates of centre ~

• $\qquad \sf( - g, - f) \: \: and \: \: (0,4)$

Here we get,

$\qquad \sf \dashrightarrow \: - g = 0$

$\qquad \sf \dashrightarrow \: g = 0$

and

$\qquad \sf \dashrightarrow \: - f = 4$

$\qquad \sf \dashrightarrow \: f = - 4$

Now, let's find the the Radius using distance formula on the given points, one of them is centre and other is point lying on circle, so distance between them is the radius.

$\qquad \sf \dashrightarrow \: r = \sqrt{( 0 - ( - 2)) {}^{2} + (4 - ( - 4)) {}^{2} }$

$\qquad \sf \dashrightarrow \: r = \sqrt{ (0 + 2) {}^{2} + (4 + 4) {}^{2} }$

$\qquad \sf \dashrightarrow \: r = \sqrt{ (2) {}^{2} + (8) {}^{2} }$

$\qquad \sf \dashrightarrow \: r = \sqrt{ 4+ 64 }$

$\qquad \sf \dashrightarrow \: r = \sqrt{ 68 }$

$\qquad \sf \dashrightarrow \: r = 2\sqrt{ 17 }$

Now, use the the following equation to find c of the standard equation

$\qquad \sf \dashrightarrow \: r = \sqrt{g {}^{2} + {f}^{2} - c}$

$\qquad \sf \dashrightarrow \: 2 \sqrt{17} = \sqrt{ {0}^{2} + {4}^{2} - c}$

squaring both sides :

$\qquad \sf \dashrightarrow \: 68 = {0}^{2} + {4}^{2} - c$

$\qquad \sf \dashrightarrow \: 68 = 8 - c$

$\qquad \sf \dashrightarrow \: - c = 68 - 8$

$\qquad \sf \dashrightarrow \: c = - 60$

Therefore, we got the standard equation of circle as ~

$\qquad \sf \dashrightarrow \: {x}^{2} + {y}^{2} + 2(0)x + 2( - 4)y + ( - 60) = 0$

$\qquad \sf \dashrightarrow \: {x}^{2} + {y}^{2} - 8y - 60 = 0$

I Hope it helps ~