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

1 Answer
Jul 9, 2016

Answer:

#(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"#