Question

A circle is drawn with AB as the diameter whose endpoints are $A(2,4)$ and $B(4,12)$. If another circle with diameter one-third of the above circle is drawn with the same centre, what are the points that the circle cuts AB?

• A. $(6,\frac{28}{3})$ and $(6,\frac{20}{3})$
• B. $(10,\frac{28}{3})$ and $(6,\frac{20}{3})$
• C. $(10,\frac{28}{3})$ and $(10,\frac{20}{3})$
• D. $(10,\frac{20}{3})$ and $(6,\frac{20}{3})$

Answer 1

The correct option is B

$(10,\frac{28}{3})$ and $(6,\frac{20}{3})$

Given, diameter AB has the endpoints as $A(2,4)$ and $B(4,12)$.

We shall first find the centre of this circle.

We know that the centre of the circle '$O(x,y)$' is the mid-point of AB.

Hence by mid-point formula, we have,

$(x,y)$ = $(\frac{2+14}{2},\frac{4+12}{2})$ = $(8,8)$

Now, if another circle with diameter one-third of the above circle is drawn with the same centre, then OP:PB = 1:2

Then by section formula, we get,

$x$-coordinate of P = $8+\frac{1}{3}(14-8)$ = $10$

$y$-coordinate of P = $8+\frac{1}{3}(12-8)$ = $\frac{28}{3}$

Therefore coordinates of P are $(10,\frac{28}{3})$

Since coordinates of P are $(10,\frac{28}{3})$ and $(8,8)$ is the centre of the circle, we must have coordinates of point Q as $(6,\frac{20}{3})$.( By mid-point formula)