Equations of two diameters of a circle are 2x – 3y = 5 and 3x – 4y = 7. The line joining the points and intersects the circle at only one point . Then is equal to __________. [2024]

Question

Equations of two diameters of a circle are 2x – 3y = 5 and 3x – 4y = 7. The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then $17 β - α$ is equal to __________. [2024]

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(2)

Solving 2x – 3y = 5 and 3x – 4y = 7, we get x = 1 and y = –1

Now, equation of tangent joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ is

$y + 4 = \frac{3 + 4}{\frac{- 1}{7} + \frac{22}{7}} (x + \frac{22}{7}) \Rightarrow 3 (y + 4) = 7 (x + \frac{22}{7})$

$\Rightarrow 7 x - 3 y + 10 = 0$

$α, β$ be the foot of perpendicular from point (1, –1) to the line 7x – 3y + 10 = 0.

$∴ \frac{α - 1}{7} = \frac{β + 1}{- 3} = \frac{- (7 (1) - 3 (- 1) + 10)}{{(- 3)}^{2} + 7^{2}} = - \frac{20}{58} = \frac{- 10}{29}$

$∴ α = \frac{- 70}{29} + 1 = \frac{- 41}{29}, β = \frac{30}{29} - 1 = \frac{1}{29}$

$∴ 17 β - α = \frac{17}{29} + \frac{41}{29} = \frac{58}{29} = 2$ .

Solution

Q.
Equations of two diameters of a circle are 2x – 3y = 5 and 3x – 4y = 7. The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then $17 β - α$ is equal to __________. [2024]

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Solution

Q. Equations of two diameters of a circle are 2x – 3y = 5 and 3x – 4y = 7. The line joining the points (–227,–4) and (–17,3) intersects the circle at only one point P(α,β). Then 17β–α is equal to __________. [2024]

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Q.
Equations of two diameters of a circle are 2x – 3y = 5 and 3x – 4y = 7. The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then $17 β - α$ is equal to __________. [2024]