Question

Through a given point $P(5,2)$ secants are drawn to cut the circle $x^2+y^2=25$ at points $A_1\left(B_1\right),A_2\left(B_2\right),A_3\left(B_3\right),A_4\left(B_4\right),A_5\left(B_5\right)$ such that $P{A}_1+P{B}_1=5,P{A}_2+P{B}_2=6,P{A}_3+P{B}_3=7,P{A}_4+P{B}_4=8,P{A}_5+P{B}_5=9$. Find the value of ${\sum }_{i=1}^{5}P{A}_i^2+{\sum }_{i=1}^{5}P{B}_i^2$
[Note: $A_r\left(B_r\right)$ denotes that the line passing through $P(5,2)$ meets the circle $x^2+y^2=25$ at two points $A_r$ and $B_r$

Answer 1

Using,

$A^2+B^2=(A+B{)}^2-2AB$

$\Rightarrow$ $P_{A1}^2+P_{A2}^2+P_{A3}^2+P_{A4}^2+P_{A5}^2+P_{B1}^2+P_{B2}^2+P_{B3}^2+P_{B4}^2+P_{B5}^2$

$\Rightarrow (P_{A1}+P_{B1}{)}^2-2P_{A1}{P}_{B1}+$$(P_{A2}+P_{B2}{)}^2-2P_{A2}{P}_{B2}+$$(P_{A3}+P_{B3}{)}^2-2P_{A3}{P}_{B3}+$

$(P_{A4}+P_{B4}{)}^2-2P_{A4}{P}_{B4}+$ $(P_{A5}+P_{B5}{)}^2-2P_{A5}{P}_{B5}$

$\Rightarrow 25+36+49+64+81-(2\times 5\times P{T}^2)$ $\Rightarrow$ $Since$ $we$ $know$ $[P_A{P}_B=P{T}^2]$

$\Rightarrow 255-10P{T}^2$ $------\left(1\right)$

Also, Using $distanceformula$ $\Rightarrow$ $PT=\sqrt{(5-5{)}^2+(2-0{)}^2}$

$\Rightarrow PT=2$

$\Rightarrow P{T}^2=4$ $-----\left(2\right)$

Using (2) and (1)

$\Rightarrow 255-10\times 4=215$