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Let $A(-1,0),B(3,0)$ and PQ be any line passing through (4,1).The range of the slope of PQ for which there are two points on PQ at which AB subtends a right angle is $(\lambda_1,\lambda_2)$,then what is $\lambda_1+\lambda_2$.

My attempt:Let equation of PQ be $y-1=m(x-4)$.Let points on PQ at which AB subtends a right angle are $C(x_1,y_1)$and$D(x_2,y_2)$.
$\Rightarrow$$y_1-1=m(x_1-4)$ and $y_2-1=m(x_2-4)$.I could not think further steps to solve the problem.Can someone help me in this question?

Vinod Kumar Punia
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1 Answers1

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HINT

The location of the points at which $AB$ substends a right angle is a $\color{red}{\text{circle}}$ of radius $2$ centered at $\color{red}{(1,0)}$ as shown below:

enter image description here

In the figure the black line through $P$ has two intersection point (green and blue) with the $\color{red}{\text{circle}}$ at the $\color{blue}{\text{point}}$ and at the $\color{green}{\text{point}}$.

The range of the straight lines form $P$ which intersect the $\color{red}{\text{circle}}$ are between the two white tangent lines. The tangent lines go through $P$ and the the intersection points of the white circle centered at $(3.5,0.5)$ and going through the center of the $\color{red}{\text{circle}}$. The center of the white circle is the midpoint of the interval joining $P$ and the center of the $\color{red}{\text{circle}}$.

zoli
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  • ,Sir,we need to find slope of $PT and PT'$ ,for that we need to find coordinates of $T$ and $T'$.Can we find them or otherwise can we find slope of $PT and PT'$. – Vinod Kumar Punia Aug 01 '15 at 14:36
  • We have two circles: $(x-1)^2+y^2=4$ (the red circle) and $(x-2.5)^2+(y-0.5)^2=\frac{10}4$ (the white circle). $T$ and $T'$ are the intersection points of these two circles. – zoli Aug 02 '15 at 09:50