Question

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

Answer 1

`▪▪▪▪▪▪▪▪▪▪▪▪▪  {\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 = √( 4+ 64 )`

`\qquad \sf  \dashrightarrow \: r = √( 68 )`

`\qquad \sf  \dashrightarrow \: r = 2√( 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 √(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 ~