Let y = x + 2 , 4 y = 3 x + 6 and 3 y = 4 x + 1 be three tangent lines to the circle ( x - h ) 2 + ( y - k ) 2 = r 2 . Then h + k is equal to [2023]

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Solution

Q.
Let $y = x + 2, 4 y = 3 x + 6$ and $3 y = 4 x + 1$ be three tangent lines to the circle ${(x - h)}^{2} + {(y - k)}^{2} = r^{2} .$ Then $h + k$ is equal to [2023]

1 5

2 $5 (1 + \sqrt{2})$

3 6

4 $5 \sqrt{2}$

Ans.
(1)

We have equation of circle ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

We know that the perpendicular distance from centre to the tangent is equal to radius of the circle.

$L_{1} : x - y + 2 = 0;$ $L_{2} : 3 x - 4 y + 6 = 0;$ $L_{3} : 4 x - 3 y + 1 = 0$

So, $r = \frac{h - k + 2}{\sqrt{2}} = \frac{3 h - 4 k + 6}{\sqrt{9 + 16}} = \frac{4 h - 3 k + 1}{\sqrt{16 + 9}}$

Consider, $\frac{3 h - 4 k + 6}{5} = \frac{4 h - 3 k + 1}{5}$

$\Rightarrow 3 h - 4 k + 6 = 4 h - 3 k + 1 \Rightarrow h + k = 5$

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