Question

Find the equation of the circle of radius 5 cm, whose centre lies on the y-axis
and which passes through the point (3,2).​

Answer 1

general equation of circle is (x-h)^2+(y-k)^2=r^2

here center lies on y-axis then let the center point be (0,a) in which abscissa is equal to zero

then (h,k)=(0,a)

given circle passes through the point (3,2)

then (x,y)=(3,2)

given radius of the circle 5cm , then r=5cm

substitute the above values in general equation of circle

then the equation becomes,

(3-0)^2+(2-a)^2=(5)^2

9+4-4a+(a^2)=25

(a^2)-4a-12=0

(a^2)-6a+2a-12=0

a(a-6)+2(a-6)=0

(a-6)(a+2)=0

then the values of a are 6 and -2

if a=6

then circle equation is (x-0)^2+(y-6)^2=0

x^2+y^2-12y+36=0

if a=-2

then circle equation is (x-0)^2+(y+2)^2=0

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