10-1.Circle and System of Circles
hard

કેન્દ્ર $(2,3)$ અને ત્રિજ્યા $4$ વાળું વર્તુળ રેખા $x+y=3$ ને બિંદુઓ $P$ અને $Q$ માં છેદે છે. જો $P$ અને $Q$ પાસેના સ્પર્શકો બિંદુ $S(\alpha, \beta)$ માં છેદે, તો $4 \alpha-7 \beta=....................$

A

$11$

B

$10$

C

$80$

D

$90$

(JEE MAIN-2023)

Solution

The given line is polar or $P (2, \beta)$ w.r.t. given circle

$x^2+y^2-4 x-6 y-3=0$

Chord or contact

$\alpha x+\beta y-2(x+\alpha)-3(y+\beta)-3=0$

$\Rightarrow(\alpha-2) x+(\beta-3) y-(2 \alpha+3 \beta+3)=0$

$\because$ But the equation of chord of contact is given

as : $x+y-3=0$

comparing the coefficients

$\frac{\alpha-2}{1}=\frac{\beta-3}{1}=-\left(\frac{2 \alpha+3 \beta+3}{-3}\right)$

On solving $\alpha=-6$

$\beta=-5$

Now $\quad 4 \alpha-7 \beta=11$

Standard 11
Mathematics

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