A line is drawn through a fixed point P(alpha | vedclass

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(B) The equation of any line passing through the point $P(\alpha, \beta)$ is given by $\frac{x-\alpha}{\cos 	heta} = \frac{y-\beta}{\sin 	heta} = k$

Vedclass · Accepted Answer

$\alpha^{2}+\beta^{2}-r^{2}$

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(B) The equation of any line passing through the point $P(\alpha, \beta)$ is given by $\frac{x-\alpha}{\cos 	heta} = \frac{y-\beta}{\sin 	heta} = k$,where $k$ represents the distance from $P$ to any point on the line.Any point on this line is $(\alpha + k \cos 	heta, \beta + k \sin 	heta)$.Since this point lies on the circle $x^{2} + y^{2} = r^{2}$,we have:$(\alpha + k \cos 	heta)^{2} + (\beta + k \sin 	heta)^{2} = r^{2}$Expanding this,we get:$\alpha^{2} + 2k\alpha \cos 	heta + k^{2} \cos^{2} 	heta + \beta^{2} + 2k\beta \sin 	heta + k^{2} \sin^{2} 	heta = r^{2}$$k^{2}(\cos^{2} 	heta + \sin^{2} 	heta) + 2k(\alpha \cos 	heta + \beta \sin 	heta) + (\alpha^{2} + \beta^{2} - r^{2}) = 0$$k^{2} + 2k(\alpha \cos 	heta + \beta \sin 	heta) + (\alpha^{2} + \beta^{2} - r^{2}) = 0$This is a quadratic equation in $k$. Let the roots be $k_{1}$ and $k_{2}$,which represent the lengths $PA$ and $PB$.The product of the roots $PA \cdot PB = k_{1}k_{2}$ is given by the constant term of the quadratic equation:$PA \cdot PB = \alpha^{2} + \beta^{2} - r^{2}$.

$A$ line is drawn through a fixed point $P(\alpha, \beta)$ to cut the circle $x^{2}+y^{2}=r^{2}$ at $A$ and $B$. Then $PA \cdot PB$ is equal to

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