Question

7- a. What is the radius of the circle with center (3,10) that passes through (12,12)?

Answer 1

Answer: a. Radius of circle = $\sqrt{85}$

b. The equation of this circle : $(x-3)^2+(y-10)^2=85$

Step-by-step explanation:

Given : Center of the circle = (3,10)

Circle is passing through (12,12).

a. To find the radius we apply distance formula (∵ Radius is the distance from center to any point ion the circle.)

Radius of circle = $\sqrt{(12-3)^2+(12-10)^2}$

Radius of circle = $\sqrt{(9)^2+(2)^2}=\sqrt{81+4}=\sqrt{85}$

i.e. Radius of circle = $\sqrt{85}$

b. Equation of a circle = $(x-h)^2+(y-k)^2=r^2$ , where (h,k)=Center and r=radius of the circle.

Put the values of (h,k)= (3,10) and r= $\sqrt{85}$ , we get

$(x-3)^2+(y-10)^2=(\sqrt{85})^2$

$(x-3)^2+(y-10)^2=85$

∴  The equation of this circle :$(x-3)^2+(y-10)^2=85$