Tangent L_1 equiv 3x - 4y - 8 = 0 and the cho | vedclass

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(D) Let the centre of the circle be $O(h, k)$ and radius be $r$. Since $L_1$ is a tangent,the perpendicular distance from $(h, k)$ to $3x - 4y - 8 = 0

Vedclass · Accepted Answer

$3$

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(D) Let the centre of the circle be $O(h, k)$ and radius be $r$. Since $L_1$ is a tangent,the perpendicular distance from $(h, k)$ to $3x - 4y - 8 = 0$ is equal to $r$.$\frac{|3h - 4k - 8|}{\sqrt{3^2 + (-4)^2}} = r \Rightarrow |3h - 4k - 8| = 5r$. Since $L_1$ is a tangent,$r = 2$,so $|3h - 4k - 8| = 10$.Also,the distance from $(h, k)$ to $L_2 \equiv x + y - 1 = 0$ is $\sqrt{2}$.$\frac{|h + k - 1|}{\sqrt{1^2 + 1^2}} = \sqrt{2} \Rightarrow |h + k - 1| = 2$.Given $h^2 + k^2 = 13$. Solving these equations,we find $(h, k) = (3, 2)$ or $(-2, 3)$.For $(h, k) = (3, 2)$,the midpoint $(\alpha, \beta)$ of the chord $L_2$ is the projection of the centre $(3, 2)$ on $x + y - 1 = 0$.$\frac{\alpha - 3}{1} = \frac{\beta - 2}{1} = -\frac{3 + 2 - 1}{1^2 + 1^2} = -\frac{4}{2} = -2$.$\alpha = 3 - 2 = 1, \beta = 2 - 2 = 0$. Thus,$\alpha + \beta = 1$.Given $r = 2$,we have $\alpha + \beta + r = 1 + 2 = 3$.

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