Question

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.

Answer 1

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