Question

Find an equation of the circle that satisfies the given conditions. (Give your answer in terms of x and y.) Center at the origin and passes through (8, 1)

Answer 1

Answer:

The equation of circle is $x^2+y^2=65$.

Step-by-step explanation:

It is given that the circle passes through the point (8,1) and center at the origin.

The distance between any point and the circle and center is called radius. it means radius of the given circle is the distance between (0,0) and (8,1).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula the radius of circle is

$r=\sqrt{\left(8-0\right)^2+\left(1-0\right)^2}=\sqrt{65}$

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$         .... (1)

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

The center of the circle is (0,0). So h=0 and k=0.

Substitute h=0, k=0 and $r=\sqrt{65}$ in equation (1).

$(x-0)^2+(y-0)^2=(\sqrt{65})^2$

$x^2+y^2=65$

Therefore the equation of circle is $x^2+y^2=65$.