Question

A square $ABCD$ has sides of length $4$ units, and $M$ is the midpoint of $CD$. A circle with radius $2$ and centre $M$ intersects another circle with radius $4$ and centre $A$ at the points $P$ and $D$. If $A$ is the origin, $B(4,0)$, then coordinates of $P$ can be

• A. $(-\frac{16}{5},\frac{12}{5})$
• B. $(\frac{16}{5},\frac{12}{5})$
• C. $(\frac{13}{4},\frac{12}{5})$
• D. $(\frac{16}{5},-\frac{12}{5})$

Answer 1

Answer: (B), (D)

The correct options are
B $(\frac{16}{5},\frac{12}{5})$
D $(\frac{16}{5},-\frac{12}{5})$

The coordinates of $A,B$ are $(0,0)$ and $(4,0)$ respectively.
Point $C$ can be either in first quadrant or fourth quadrant.

So, its coordinates can be $(4,4)$ or $(4,-4)$
So, accordingly $D$ can be $(0,4)$ or $(0,-4)$.
Let us assume $C$ and $D$ to be in the first quadrant.
The coordinates of mid-point $M$ of $CD$ would be $(2,4)$
So, the circle with center at $M$ and radius $2$ will be given by,
$(x-2{)}^2+(y-4{)}^2=4$
$\Rightarrow x^2+y^2-4x-8y+16=0$
The other circle with centre at $A$ and radius $4$ is given by,
$x^2+y^2=16$
Let the two circles intersect at $P$ $(x_1,y_1)$
$\Rightarrow 16-4x_1-8y_1+16=0$
$\Rightarrow x_1+2y_1=8$
On substituting the value of $x$ in the equation of the circle we get, the coordinates of $P$ are $(\frac{16}{5},\frac{12}{5})$.
If $P$ lies in fourth quadrant, then coordinates of $P$ are $(\frac{16}{5},-\frac{12}{5})$.