Question

find the equation of the circle passing through the point (7,3) having radius 3 units and whose centre lies on the line y = x-1.

Answer 1

first draw the circle then ask this

Answer 2

Answer:

$(x-4)^2+(y-3)^2=9$ or

$(x-7)^2+(y-6)^2=9$

Step-by-step explanation:

A circle passes through the point (7,3) having radius 3 units and whose centre lies on the line y=x-1.

Let us suppose centre be (h,k)

Centre lie on line.

k=h-1

using distance formula between (h,k) and (7,3) is 3

$\sqrt{(h-7)^2+(k-3)^2}=3$

Substitute k into this equation and solve for h

$(h-7)^2+(h-1-3)^2=3^2$

$h^2+49-14h+h^2+16-8h=9$

$2h^2-22h+56=0$

$h^2-11h+28=0$

$(h-7)(h-4)=0$

$h=4,7$

For h=4, k=3

For h=7, k=6

Centre: (4,3)  or  (7,6)

Equation of circle:

$(x-4)^2+(y-3)^2=9$ or

$(x-7)^2+(y-6)^2=9$

Please see the attachment for figure.