Question

If $△PAB$ is right angle at $P$, where $A(2,3)$ and $B(5,7)$, then which of the following is/are correct?

• A. Equation of circumcircle of $△PAB$ is $x^2+y^2-7x-10y+31=0$
• B. Maximum possible area of $△PAB$ is $\frac{25}{4}\text{sq. units}.$
• C. When area of $△PAB$ is $6\text{sq. units}$, then there are $4$ possible values of $P.$
• D. When area of $△PAB$ is $6\text{sq. units}$, then there are $2$ possible values of $P.$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Equation of circumcircle of $△PAB$ is $x^2+y^2-7x-10y+31=0$
B Maximum possible area of $△PAB$ is $\frac{25}{4}\text{sq. units}.$
C When area of $△PAB$ is $6\text{sq. units}$, then there are $4$ possible values of $P.$

Assuming $AB$ as a diameter of a circle, then $P$ lies on the circle and $△PAB$ is right angle at $P$


Equation of circumcircle of $△PAB$ is
$(x-2)(x-5)+(y-3)(y-7)=0\Rightarrow x^2+y^2-7x-10y+31=0$

Length of $AB$
$d=\sqrt{3^2+4^2}=5\Rightarrow r=\frac{5}{2}$

Area of $△PAB$
$=\frac{1}{2}\times PC\times 2r=PC\times r$
Maximum possible area occur when $PC=r$,
$=r^2=\frac{25}{4}\text{sq. units}$

When area of $△PAB$ is $6\text{sq. units}$, then
$6=PC\times r\Rightarrow PC=\frac{12}{5}<r$
Therefore, there are $4$ possible values of $P$ as seen by the figure, two above and two below.