Equation 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 \(P(\alpha,\beta)\).Then \(17\beta-\alpha\) is equal to.
Correct Answer: 2
Solution and Explanation
Step 1: Find the Centre of the Circle
The centre \(C\) of the circle is the intersection of the diameters \(2x - 3y = 5\) and \(3x - 4y = 7\). Solving these equations, we get \(C(1, -1)\).
Step 2: Equation of Line \(AB\)
The points \(A\left(-\frac{22}{7}, -4\right)\) and \(B\left(\frac{1}{7}, 3\right)\) lie on the line \(AB\). The equation of \(AB\) is:
\[ 7x - 3y + 10 = 0 \quad (i) \]
Step 3: Equation of Line \(CP\)
Since \(P\) lies on the circle, \(CP\) is perpendicular to \(AB\) with the equation:
\[ 3x + 7y + 4 = 0 \quad (ii) \]
Step 4: Solve for \(\alpha\) and \(\beta\)
Solving equations (i) and (ii), we find:
\[ \alpha = -\frac{41}{29}, \quad \beta = \frac{1}{29} \]
Step 5: Calculate \(17\beta - \alpha\)
\[ 17\beta - \alpha = 17 \cdot \frac{1}{29} + \frac{41}{29} = 2 \]
So, the correct answer is: 2
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