Question

What is a equation that has a center at (-5,2) with a radius of 6?

Answer 1

The given point (center of the circle) is `(-5,2)`
Let the unknown point on the circumference of the circle be `(x_1,y_1)`

`color(white)(dd`

The radius of the circle is given by `sqrt((x_1-h)^2+(y_1-k)^2)`
(its just basically the distance formula)

`color(white)(dd`

And therefore, the general equation of a circle having a center `(h,k)` and radius `r` is `(x-h)^2+(y-k)^2=r^2`.

So, in this case, since the center is `(-5,2)`
`(x-(-5))^2+(y-(2))^2=(6)^2`.

`=> (x+5)^2+(y-2)^2=36`.

`=> x^2 + 25+ 10x + y^2 +4 +4x = 36`

`=> color(red)(x^2 + y^2 + 10x + 4y - 7=0`

Answer 2

`(x+5)^2+(y-2)^2=36`

Explanation:

`"the standard form of the 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 is"`
`"the radius"`

`"here "(a,b)=(-5,2)" and " r=6`

`(x-(-5))^2+(y-2)^2=6^2`

`rArr(x+5)^2+(y-2)^2=36larrcolor(red)"equation of circle"`