Let the centre of a circle C be and its radius r 8 . Let 3 x + 4 y = 24 and 3 x - 4 y = 32 be two tangents and 4 x + 3 y = 1 be a normal to C. Then is equal to [2023]

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Solution

Q.
Let the centre of a circle C be $(α, β)$ and its radius $r < 8$ . Let $3 x + 4 y = 24$ and $3 x - 4 y = 32$ be two tangents and $4 x + 3 y = 1$ be a normal to C. Then $(α - β + r)$ is equal to [2023]

1 6

2 5

3 7

4 9

Ans.
(3)

$After solving 4 x + 3 y = 1 and 3 x - 4 y = 32, Point A is (4, - 5)$

$Centre (α, β) lies on the line 4 x + 3 y = 1$

$So, 4 α + 3 β = 1$

$∴$ $β = \frac{1 - 4 α}{3}$

$Now, distance from Centre to line 3 x - 4 y - 32 = 0$ $and 3 x + 4 y - 24 = 0 are equal.$

$So,$ $| \frac{3 α - 4 (\frac{1 - 4 α}{3}) - 32}{\sqrt{3^{2} + 4^{2}}} |$ $=$ $| \frac{3 α + 4 (\frac{1 - 4 α}{3}) - 24}{5} |$

$\Rightarrow$ $| \frac{3 α - 4 (\frac{1 - 4 α}{3}) - 32}{5} |$ $=$ $| \frac{3 α + 4 (\frac{1 - 4 α}{3}) - 24}{5} |$

$\Rightarrow$ $| \frac{9 α - 4 + 16 α - 96}{15} |$ $=$ $| \frac{9 α + 4 - 16 α - 72}{15} |$

$\Rightarrow 25 α - 100 = \pm (- 7 α - 68)$

$∴$ $α = 1 and α = \frac{28}{3}$

For $α = 1$ , Centre is $(1, - 1)$

and $radius is \sqrt{{(4 - 1)}^{2} + {(- 5 + 1)}^{2}} = 5 units$

For $α = \frac{28}{3}$ , $β = \frac{- 109}{9}, Centre$ $(\frac{28}{3}, \frac{- 109}{9})$

$∴$ $radius≃ 8.88 (rejected)$

Hence, $α = 1, β = - 1, r = 5$

$∴$ $α - β + r = 7$

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