JEE Main 2024 — Mathematics Coordinate Geometry

Given diameters of circle are $2x-3y=5$2x−3y=5 and $3x-4y=7$3x−4y=7.

We know that, the intersection point of diameters is centre.

$$\Rightarrow x=\frac{5+3y}{2}=\frac{7+4y}{3}$$⇒x=25+3y​=37+4y​

$$\Rightarrow 15+9y=14+8y$$⇒15+9y=14+8y

$$\Rightarrow y=-1$$⇒y=−1

$$\Rightarrow x=\frac{5-3}{2}=1$$⇒x=25−3​=1

So, centre of circle is $(1,-1)$(1,−1)

Equation of $AB$AB is given by,

$$y+4=\frac{3+4}{\frac{-1}{7}+\frac{22}{7}}(x+\frac{22}{7})$$y+4=7−1​+722​3+4​(x+722​)

$$\Rightarrow y+4=\frac{7}{3}(\frac{7x+22}{7})$$⇒y+4=37​(77x+22​)

$$\Rightarrow 3y+12=7x+22$$⇒3y+12=7x+22

$$\Rightarrow 7x–3y+10=0...(i)$$⇒7x–3y+10=0...(i)

Slope of $AB,{m}_{1}=\frac{7}{3}$AB,m1​=37​

So, slope of $CP,{m}_{2}=\frac{-1}{{m}_{1}}=\frac{-3}{7}$CP,m2​=m1​−1​=7−3​

Equation of $CP$CP is given by,

$$(y+1)=(\frac{-3}{7})(x-1)$$(y+1)=(7−3​)(x−1)

$$\Rightarrow 7y+7=-3x+3$$⇒7y+7=−3x+3

$$\Rightarrow 3x+7y+4=0...(ii)$$⇒3x+7y+4=0...(ii)

Solving $(i)\text{and}(ii)$(i)and(ii),

$$\Rightarrow x=\frac{3y-10}{7}=\frac{-7y-4}{3}$$⇒x=73y−10​=3−7y−4​

$$\Rightarrow 9y-30=-49y-28$$⇒9y−30=−49y−28

$$\Rightarrow 58y=2$$⇒58y=2

$$\Rightarrow y=\frac{1}{29}$$⇒y=291​

$$\Rightarrow x=\frac{3(\frac{1}{29})-10}{7}=\frac{3-290}{7\times 29}$$⇒x=73(291​)−10​=7×293−290​

$$\Rightarrow x=\frac{-41}{29}$$⇒x=29−41​

So, $\alpha =\frac{-41}{29},\beta =\frac{1}{29}$α=29−41​,β=291​

$$\Rightarrow 17\beta -\alpha =\frac{17}{29}+\frac{41}{29}$$⇒17β−α=2917​+2941​

$$\Rightarrow 17\beta -\alpha =2$$⇒17β−α=2