The point of intersection of the tangents dra | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 4x - 6y + 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2, f = -3, c = 4$. The ce

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$\left(-\frac{5}{2}, \frac{21}{4}\right)$

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(B) The equation of the circle is $x^2 + y^2 - 4x - 6y + 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2, f = -3, c = 4$. The center of the circle is $(h, k) = (-g, -f) = (2, 3)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - 4} = 3$. The line is $L: 2x - y + 3 = 0$. The point of intersection $(x_1, y_1)$ of the tangents at the points of intersection of the line $lx + my + n = 0$ and the circle $(x-h)^2 + (y-k)^2 = r^2$ is given by the formula: $x_1 = h - \frac{r^2 l}{lh + mk + n}$ and $y_1 = k - \frac{r^2 m}{lh + mk + n}$. Here,$l = 2, m = -1, n = 3, h = 2, k = 3, r^2 = 9$. Denominator $D = lh + mk + n = 2(2) + (-1)(3) + 3 = 4 - 3 + 3 = 4$. $x_1 = 2 - \frac{9(2)}{4} = 2 - \frac{18}{4} = 2 - 4.5 = -2.5 = -\frac{5}{2}$. $y_1 = 3 - \frac{9(-1)}{4} = 3 + \frac{9}{4} = \frac{12 + 9}{4} = \frac{21}{4}$. Thus,the point of intersection is $\left(-\frac{5}{2}, \frac{21}{4}\right)$.

The point of intersection of the tangents drawn at the points where the line $2x - y + 3 = 0$ meets the circle $x^2 + y^2 - 4x - 6y + 4 = 0$ is

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