Question

Equation 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 \(\mathrm{P}(\alpha, \beta)\). Then \(17 \beta-\alpha\) is equal to _____.

Answer 1

Correct answer: 2

Centre of circle is (1, -1)

Equation of \(\mathrm{AB}\) is \(7 \mathrm{x}-3 \mathrm{y}+10=0 \quad ....(i)\)

Equation of \(\mathrm{CP}\) is \(3 \mathrm{x}+7 \mathrm{y}+4=0 \quad ....(ii)\)

Solving (i) and (ii)

\(\alpha=\frac{-41}{29}, \beta=\frac{1}{29}\)

\( \therefore 17 \beta-\alpha=2\)

Answer 2

2 x − 3 y = 5 and 3 x − 4 y = 7 are two lines that passes through centre of the circle.

Therefore, their intersection point i.e. (1,-1) is centre of the circle.

Given,

line joining the points ( − 22/7 , − 4 )   and   ( − 1/7 , 3 ) intersects the circle at only one point P ( α , β )

With two point formula, we get

y = 7/3x + 10/3

 7/3x - y + 10/3=0

7x - 3y +10 = 0 

Comparing above equation with ax+by+c=0 we get

a=7

b=-3

c=10

(x1,y1)=(1,-1)

Using the formula for finding the perpendicular distance of a point from a line we get

d = (|7+3+10|)/\({\sqrt{7^2 + (-3)^2}}\)

= 20/\({\sqrt{58}}\)

= 6.28 units

Using distance formula

d = \({\sqrt{(α-1)^2 + (β+1)^2 }}\)    --------(i)

7α -3β +10 = 0 --------(ii)

By equation i and ii,

α = 0.8

β= 1.86 

OR

α = -3.6

β= 5.067

17 β - α= 29.76 or 89.73