Question

The circle x^{2}+y^{2}=4x2+y2=4 cuts the line joining the points A(1,0)A(1,0) and B(3,4)B(3,4) in two points PP and
Q let \frac{B P}{P A}=\alpha,PABP​=α, and \frac{B Q}{Q A}=\betaQABQ​=β then \alphaα and \betaβ are roots of the equation​

Answer 1

Given: the circle $x^{2}+y^{2}=4$ cuts the line joining the points $A\:(1,\:0)$ and $B\:(3,\:4)$ in two points $P$ and $Q$

To find: let $\frac{BP}{PA}=\alpha$ and $\frac{BQ}{QA}=\beta$; then $\alpha$ and $\beta$ are the roots of which equation

Solution:

The given circle is $\x^{2}+y^{2}=4\:\:...(1)$

The straight line passing through the points $A\:(1,\:0)$ and $B\:(3,\:4)$ is given by

$\quad \frac{y-0}{0-4}=\frac{x-1}{1-3}$

$\Rightarrow \frac{y}{-4}=\frac{x-1}{-2}$

$\Rightarrow -2y=-4x+4$

$\Rightarrow 2y=4x-4$

$\Rightarrow y=2x-2$

Now substituting the value of $y$ in $(1)$ no. equation, we get

$\quad x^{2}+(2x-2)^{2}=4$

$\Rightarrow x^{2}+4x^{2}-8x+4=4$

$\Rightarrow 5x^{2}-8x=0$

$\Rightarrow x\:(5x-8)=0$

This gives: $x=0,\:\frac{8}{5}$

• When $x=0$, we get $y=-2$

• When $x=\frac{8}{5}$, we get $y=\frac{6}{5}$

So we have the following points:

• $A\:(1,\:0)$
• $B\:(3,\:4)$
• $P\:(0,\:-2)$
• $Q\:(\frac{8}{5},\:\frac{6}{5})$

Now we find the distances $BP$, $PA$, $BQ$ and $QA$ using the formuma for distance between two points:

• $BP=\sqrt{(3-0)^{2}+(4+2)^{2}}$ units
• $=\sqrt{9+16}$ units
• $=\sqrt{25}$ units
• $=5$ units

• $PA=\sqrt{(1-0)^{2}+(0+2)^{2}}$ units
• $=\sqrt{1+4}$ units
• $=\sqrt{5}$ units

• $BQ=\sqrt{(3-\frac{8}{5})^{2}+(4-\frac{6}{5})^{2}}$ units
• $=\sqrt{\frac{49}{25}+\frac{196}{25}}$ units
• $=\sqrt{\frac{245}{25}}$ units
• $=\sqrt{\frac{49}{5}}$ units
• $=\frac{7}{\sqrt{5}}$ units

• $QA=\sqrt{(\frac{8}{5}-1)^{2}+(\frac{6}{5}-0)^{2}}$ units
• $=\sqrt{\frac{9}{25}+\frac{36}{25}}$ units
• $=\sqrt{\frac{45}{25}}$ units
• $=\sqrt{\frac{9}{5}}$ units
• $=\frac{3}{\sqrt{5}}$ units

Now, $\alpha=\frac{BP}{PA}=\frac{5}{\sqrt{5}}=\sqrt{5}$

and $\beta=\frac{BQ}{QA}=\frac{7}{3}$

We can find that $\alpha$ and $\beta$ are the roots of the equation:

$\quad (x-\sqrt{5})(x-\frac{7}{3})=0$

$\Rightarrow x^{2}-(\sqrt{5}+\frac{7}{3})\:x+\frac{7}{3}\sqrt{5}=0$

$\Rightarrow 3x^{2}-(7+3\sqrt{5})\:x+7\sqrt{5}=0$

Answer: $3x^{2}-(7+3\sqrt{5})\:x+7\sqrt{5}=0$