A circle with centre (2, 3) and radius 4 inte | vedclass

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(A) The equation of the circle with centre $(2, 3)$ and radius $r = 4$ is $(x - 2)^2 + (y - 3)^2 = 4^2$,which simplifies to $x^2 + y^2 - 4x - 6y - 3 =

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$11$

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(A) The equation of the circle with centre $(2, 3)$ and radius $r = 4$ is $(x - 2)^2 + (y - 3)^2 = 4^2$,which simplifies to $x^2 + y^2 - 4x - 6y - 3 = 0$.The line $x + y = 3$ is the chord of contact for the point $S(\alpha, \beta)$ with respect to the circle.The equation of the chord of contact for the point $(\alpha, \beta)$ is given by $T = 0$,which is $x\alpha + y\beta - 2(x + \alpha) - 3(y + \beta) - 3 = 0$.Rearranging the terms,we get $(\alpha - 2)x + (\beta - 3)y - (2\alpha + 3\beta + 3) = 0$.Comparing this with the given chord equation $x + y - 3 = 0$,we have:$\frac{\alpha - 2}{1} = \frac{\beta - 3}{1} = \frac{2\alpha + 3\beta + 3}{3}$.From $\frac{\alpha - 2}{1} = \frac{\beta - 3}{1}$,we get $\alpha - 2 = \beta - 3$,so $\beta = \alpha + 1$.Substituting $\beta = \alpha + 1$ into $\alpha - 2 = \frac{2\alpha + 3(\alpha + 1) + 3}{3}$:$3(\alpha - 2) = 2\alpha + 3\alpha + 3 + 3$$3\alpha - 6 = 5\alpha + 6$$-2\alpha = 12 \Rightarrow \alpha = -6$.Then $\beta = -6 + 1 = -5$.Finally,$4\alpha - 7\beta = 4(-6) - 7(-5) = -24 + 35 = 11$.

$A$ circle with centre $(2, 3)$ and radius $4$ intersects the line $x + y = 3$ at the points $P$ and $Q$. If the tangents at $P$ and $Q$ intersect at the point $S(\alpha, \beta)$,then $4 \alpha - 7 \beta$ is equal to $........$.

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