Question

If the coordinates of the circumcentre of triangle PQR with vertices P, Q and R as (4, m), (7, 2) and (3, n) respectively are (5, 1), then the value of $(m^2-n^2)$ is (m, n > 0)

• A. -3
• B. 5
• C. 8
• D. 6

Answer 1

The correct option is B 5

Let C(5, 1) be the circumcentre of the ΔPQR.



Now, PC = QC = RC

$\Rightarrow P{C}^2=Q{C}^2=R{C}^2$

$\Rightarrow (5-4{)}^2+(1-m{)}^2=(5-7{)}^2+(1-2{)}^2=(5-3{)}^2+(1-n{)}^2$

$\Rightarrow 1+1+m^2-2m=(-2{)}^2+(-1{)}^2=(2{)}^2+1+n^2-2n$

$\Rightarrow 2+m^2-2m=4+1=4+1+n^2-2n$

$\Rightarrow m^2-2m+2=5=n^2-2n+5$

$\Rightarrow m^2-2m+2=5and{n}^2-2n+5=5$

$\Rightarrow m^2-2m-3=0and{n}^2-2n=0$

Now, consider $m^2-2m-3=0$

$\begin{array}{cc}D& =b^2-4ac\\ & =(-2{)}^2-4\times 1\times (-3)\\ & =4+12\\ & =16\end{array}$

Therefore, the roots of the equation are $\frac{-b\pm \sqrt{D}}{2a}=\frac{2\pm 4}{2}=3,-1.$

$\Rightarrow m=3orm=-1$

Also, consider $n^2-2n=0.$

$\Rightarrow n(n-2)=0$

⇒ n = 0 or n = 2

Given that m, n > 0.

∴ m = 3 and n = 2

Thus, $m^2-n^2=(3{)}^2-(2{)}^2=9-4=5$

Hence, the correct answer is option (b).