Question

(a) Check whether the circle with centre at point $(2,4)$ and radius $5$ units passes through the point $(2,0)$.
(b) Write the co-ordinates of the points at which this circle cuts the x-axis.

Answer 1

The general equation of a circle with centre $(h,k)$ and radius ${}^{\prime }{r}^{\prime }$ units is $(x-h{)}^2+(y-k{)}^2=r^2$
Given: Centre of circle is $(2,4)$ and radius $=5units$
Thus, the equation of a given circle is
$(x-2{)}^2+(y-4{)}^2=5^2$
$\Rightarrow x^2-4x+4+y^2-8y+16=25$
$\Rightarrow x^2+y^2-4x-8y+20=25$
$\Rightarrow x^2+y^2-4x-8y-5=0....\left(i\right)$
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. $=2^2+0^2-4\times 2-8\times 0-5$
$=4+0-8-0-5$
$=-9\ne 0$
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
$x^2+0^2-4x-8\left(0\right)-5=0$
$\Rightarrow x^2-4x-5=0$
$\Rightarrow x^2-5x+x-5=0$
$\Rightarrow x(x-5)+1(x-5)=0$
$\Rightarrow (x-5)(x+1)=0$
$\Rightarrow x=-1$ or $x=5$
Thus, the co-ordinates of the points at which the circle cuts the x-axis are $(-1,0)$ and $(5,0)$.