Question

Acircle has radius 3 units and its centrer
lies on the line y=r-1. If it passes
through the point (7,3), its equation is​

Answer 1

Answer:

Step-by-step explanation:

radius = 3

center lies on the line y=r-1

substitute r as 3

y =3-1=2

center point is (0,2) =h,k

and the circle passes through the point (7,3)

Equation of the circle passes through point (7,3) ,Center(0,2) and radius 3 is $(x-h)^2 + (y-k)^2 = r^2$

(x-0)^2 + (y-2)^2 = 3^2

x^2 + (y-2)^2 = 9

x^2 + y^2+2^2 - 2(y)(2) = 9

x^2 + y^2 -4y +4 -9 =0

x^2+y^2 -4y -5 =0

Hope this will help you

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