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posledela
1 year ago
6

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

Mathematics
1 answer:
ser-zykov [4K]1 year ago
6 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

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 ~

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