Equations of two diameters of a circle are 2 x-3 y=5 and 3 x-4 y=7.

Question

Equations of two diameters of a circle are $2 x-3 y=5$ and $3 x-4 y=7$. The line joining the points $$\left(-\frac{22}{7},-4\right)$$ and $$\left(-\frac{1}{7}, 3\right)$$ intersects the circle at only one point $$P(\alpha, \beta)$$. The...

Answer: 2

Accepted Answer

Centre of circle is $(1,-1)$

Equation of $AB$ is $7x-3y+10=0\ldots$ (i)

Equation of $CP$ is $3x+7y+4=0\ldots$ (ii)

Solving (i) and (ii)

$\alpha =\frac{-41}{29},\beta =\frac{1}{29}\, 	herefore 17\beta -\alpha =2$

Solutions · 1