Question

How do find the standard equation of the circle with center (3,6) and passes through (-1,4)?

Answer 1

`(x-3)^2+(y-6)^2=20`

Explanation:

Standard equation of a circle with center `(h,k)` and radius `r` is

`(x-h)^2+(y-k)^2=r^2`

As center is `(3,6)` and circle passes through `(-1,4)`, the radius is distance between `(3,6)` and `(-1,4)` i.e.

`r=sqrt((3-(-1))^2+(6-4)^2)=sqrt(4^2+2^2)=sqrt(16+4)=sqrt20`

Hence standard equation of circle is

`(x-3)^2+(y-6)^2=(sqrt20)^2`

or `(x-3)^2+(y-6)^2=20`

graph{((x-3)^2+(y-6)^2-20)((x-3)^2+(y-6)^2-0.02)((x+1)^2+(y-4)^2-0.02)=0 [-8.13, 11.87, 0.64, 10.64]}

Answer 2

`(x-3)^2+(y-6)^2=20`

Explanation:

`"the standard equation of a circle is "`

`color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))`

`"where "(a,b)" are the coordinates of the centre and r the"`
`"radius"`

`"the radius is the distance from the centre to the "`
`"point "(-1,4)`

`"calculate r using the "color(blue)"distance formula"`

`•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`"with "(x_1,y_1)=(3.6)" and "(x_2,y_2)=(-1,4)`

`rArrr=sqrt((-1-3)^2+(4-6)^2)=sqrt20`

`"here "(a,b)=(3,6)" and "r^2=(sqrt20)^2=20`

`rArr(x-3)^2+(y-6)^2=20larrcolor(red)" in standard form"`