Question

How do you write an equation for a circle with point (3, 4) lies on a circle whose centre is at (-1, 2)?

Answer 1

`(x+1)^2+(y-2)^2=20`

Explanation:

The standard form of the equation of a circle is

`color(red)(|bar(ul(color(white)(a/a)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(a/a)|)))`
where (a ,b) are the coordinates of the centre and r, the radius.

We are given the centre but have to find the radius.

Since we are given a point on the circle then the distance from this point to the centre is the `color(blue)"radius"`

To calculate the radius use the `color(blue)"distance formula"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(a/a)|)))`
where `(x_1,y_1)" and " (x_2,y_2)" are 2 coordinate points"`

Here the 2 points are (3 ,4) and (-1 ,2)

let `(x_1,y_1)=(3,4)" and " (x_2,y_2)=(-1,2)`

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

Substitute a = -1 , b = 2 and r`=sqrt20` into the equation of the circle.

`(x-(-1))^2+(y-2)^2=(sqrt20)^2`

`rArr(x+1)^2+(y-2)^2=20" is the equation of the circle"`