Question

Write the equation of a circle with center (6,4) that passes through the coordinate (2,1). In your final answer, include all of your calculations.

Answer 1

r^2=(x-6)^2+(y-4)^2

r^2=(6-2)^2+(4-1)^2, r^2=16+9=25

(x-6)^2+(y-4)^2=25

Answer 2

`(x-6)^2+(y-4)^2=25`

Explanation:

It is given that,

Center of the circle = (6,4)

Circle passes through the point = (2,1)

Radius of the circle is distance between center (6,4) and point on he circle (2,1).

`r=√((x_2-x_1)^2+(y_2-y_1)^2)`

`r=√((2-6)^2+(1-4)^2)`

`r=√((-4)^2+(-3)^2)`

`r=√(16+9)`

`r=√(25)`

`r=5`

Standard form of a circle is

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

where, r is radius and (h,k) is center.

Substitute h=6, k=4 and r=5 in the above equation.

`(x-6)^2+(y-4)^2=5^2`

`(x-6)^2+(y-4)^2=25`

Therefore, the equation of circle is
`(x-6)^2+(y-4)^2=25`.