Question

Consider the points $A(4,3)$ and $B(0,1)$. If a circle is drawn with AB as diameter, then the coordinates of the points where this circle intersects the $x$-axis are

• A. $(-3,0)$ and $(2,0)$.
• B. $(2,0)$ and $(-1,0)$.
• C. $(3,0)$ and $(-1,0)$.
• D. $(3,0)$ and $(1,0)$

Answer 1

The correct option is D

$(3,0)$ and $(1,0)$

We shall first find the equation of the circle whose diameter has the end points $A(4,3)$ and $B(0,1)$.

Equation of a circle when the end points $(x_1,y_1)$ and $(x_2,y_2)$ of its diameter are given, is given by,

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

Then $(x-4)(x-0)+(y-3)(y-1)=0$

$⟹x^2-4x+y^2-4y+3=0$

$⟹x^2+y^2-4x-4y+3=0$

Let the circle touches the $x$-axis at $(x,0)$.

Then, from the equation of the circle obtained above, we have,

$x^2+0^2-4x-4\left(0\right)+3=0$

i.e. $x^2-4x+3=0$

This is the equation to determine the $x$-coordinates of the points where this circle intersects $x$-axis.

Then since $x^2-4x+3=0$, we must have, $(x-3)(x-1)=0$, which implies $x=3$ or $x=1$.

Therefore the circle touches the $x$-axis at $(3,0)$ and $(1,0)$.