The general equation of a circle with centre (h,k) and radius ′r′ units is (x−h)2+(y−k)2=r2
Given: Centre of circle is (2,4) and radius =5 units
Thus, the equation of a given circle is
(x−2)2+(y−4)2=52
⇒x2−4x+4+y2−8y+16=25
⇒x2+y2−4x−8y+20=25
⇒x2+y2−4x−8y−5=0....(i)
Now, to check whether the point (2,0) passes through the circle or not, substitute x=2 and y=0 in the L.H.S. of equation (i).
Thus, we have L.H.S. =22+02−4×2−8×0−5
=4+0−8−0−5
=−9≠0
Hence, the point (2,0) does not lie on the circle.
(b) The circle cuts the x-axis at y=0
Substituting the value y=0 in equation (i), we get
x2+02−4x−8(0)−5=0
⇒x2−4x−5=0
⇒x2−5x+x−5=0
⇒x(x−5)+1(x−5)=0
⇒(x−5)(x+1)=0
⇒x=−1 or x=5
Thus, the co-ordinates of the points at which the circle cuts the x-axis are (−1,0) and (5,0).