If P is a point on the circle x^2+y^2=4, Q is a point on the straight line 5 x+y+2=0 and x…

Question

If $P$ is a point on the circle $x^2+y^2=4, Q$ is a point on the straight line $5 x+y+2=0$ and $x-y+1=0$ is the perpendicular bisector of PQ , then 13 times the sum of abscissa of all such points P is $\_\_\_\_$ .

Answer: 2

Accepted Answer

$P({x}_{1}{y}_{1})$ and point $Q({x}_{2},{y}_{2})$

Mid point of $PQM=(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})$

Substitute M into $x-y+1=0$

${x}_{1}+{x}_{2}-{y}_{1}-{y}_{2}+2=0..(i)$

Slope of PQ is Perpendicular to slope of bisector line So, slope of $PQ=-1$

${y}_{2}={x}_{1}-{x}_{2}+{y}_{1}\ldots ..(ii)$

$Q({x}_{2},{y}_{2})$ lie on $5x+y+2=0$

So, $5{x}_{2}+{y}_{2}+2=0\ldots \ldots$. (iii)

Substitute (iii) in (i)

${x}_{2}=\frac{-{x}_{1}-{y}_{1}-2}{4}\ldots .(iv)$

Substitute (iii) in (ii)

${x}_{2}={y}_{1}-1\ldots ..(v)$

From (iv) and (v)

${x}_{1}=2-5y$

$({x}_{1},{y}_{1})$ lie on circle

${x}_{2}^{1}+{y}_{2}^{1}=4$

Pt ${x}_{1}=2-5{y}_{1}$

${y}_{1}=0,-\frac{10}{13}$

So, ${x}_{1}=2,\frac{-24}{13}$

So, $2+(-\frac{24}{13})=\frac{2}{13}$

So, $13	imes \frac{2}{13}=2$

Solutions · 1