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

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(B) Let the equation of the line passing through the point $P(\alpha, \beta)$ be $\frac{x - \alpha}{\cos 	heta} = \frac{y - \beta}{\sin 	heta} = k$,

Vedclass · Accepted Answer

$\alpha^2   \beta^2 - r^2$

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(B) Let the equation of the line passing through the point $P(\alpha, \beta)$ be $\frac{x - \alpha}{\cos 	heta} = \frac{y - \beta}{\sin 	heta} = k$,where $k$ is the distance of any point $(x, y)$ on the line from the point $P(\alpha, \beta)$.Any point on this line is given by $(\alpha + k \cos 	heta, \beta + k \sin 	heta)$.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$$\alpha^2 + k^2 \cos^2 	heta + 2 \alpha k \cos 	heta + \beta^2 + k^2 \sin^2 	heta + 2 \beta k \sin 	heta = r^2$$k^2 + 2k(\alpha \cos 	heta + \beta \sin 	heta) + (\alpha^2 + \beta^2 - r^2) = 0$This is a quadratic equation in $k$. Let its roots be $k_1$ and $k_2$,which represent the distances $PA$ and $PB$ respectively.The product of the roots $k_1 k_2 = PA \cdot PB$ is given by the constant term of the quadratic equation:$PA \cdot PB = \alpha^2 + \beta^2 - r^2$.Alternatively,by the power of a point theorem,$PA \cdot PB = PT^2$,where $PT$ is the length of the tangent from $P$ to the circle,which is $\sqrt{\alpha^2 + \beta^2 - r^2}$. Thus,$PA \cdot PB = \alpha^2 + \beta^2 - r^2$.

$A$ 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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