Question

Find the eqn of a circle with radius 5 whose center lies on x axis and passes through the point 2,3

Answer 1

heyyya your answer




GIVEN: r = 5

c(2,3)

centre lie on x-axis

i.e c(a,0)

A.T.Q

eq. (2-h)2 + (3-k)2 = 25 ----------------(1)

= (2-a)2 + y2 =25 -----------------------(2)

= 4 - 4a + a2 = 25

= a2 - 4a - 21 =0

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

therefore a=6 or a= -2

hence

eq. a=6

x2 + y2 - 12x + 11= 0 ---------------ans

And a= -2

x2 + y2 + 4X -21 =0 -----------------ans

Answer 2

The equation of circle will be in the form
(x-k)^2 + (y-m)^2 = r^2
where (k,m) is centre,r is radius
the circle passes through point (2,3)
as centre lies on x axix its y co ordinate must be equal to 0..(m=0)
substituting these two conditions in the quation,we get
(2-k)^2 +(3-0)^2= 5^2
(2-k)^2=16
2-k= 4 or 2-k =-4
k= -2 or k=6
(k,m) = (-2,0) or (k,m) = (6,0)
therefore equation
(x-k)^2+(y-m)^2 = r^2
if (k,m) = (-2,0),r =5
equation= (x+2)^2+y^2= 5^2
if (k,m)= (6,0)
equation = (x-6)^2+ y^2 = 5^2
hope it helps