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
Approach Solution - 1
Step 1: Find the center of the circle.
The diameters intersect at the center of the circle. So, we solve the two linear equations: \[ 2x - 3y = 5 \quad \text{and} \quad 3x - 4y = 7 \]
Multiply the first equation by 3 and the second by 2 to eliminate \( x \): \[ 6x - 9y = 15 \\ 6x - 8y = 14 \] Subtracting: \[ (6x - 9y) - (6x - 8y) = 15 - 14 \] \[ -y = 1 \implies y = -1 \]
Substitute \( y = -1 \) in \( 2x - 3y = 5 \):
\[ 2x - 3(-1) = 5 \implies 2x + 3 = 5 \implies 2x = 2 \implies x = 1 \]
Therefore, the center of the circle is: \[ C(1, -1) \]
Step 2: Equation of the line joining given points
Let the two points be: \[ A\left(-\frac{22}{7}, -4\right) \quad \text{and} \quad B\left(-\frac{1}{7}, 3\right) \]
Slope of line \( AB \): \[ m = \frac{3 - (-4)}{-\frac{1}{7} - (-\frac{22}{7})} = \frac{7}{\frac{21}{7}} = \frac{7}{3} \]
Equation of line using point \( A(x_1, y_1) \): \[ y + 4 = \frac{7}{3}\left(x + \frac{22}{7}\right) \]
Simplify: \[ 3(y + 4) = 7\left(x + \frac{22}{7}\right) \] \[ 3y + 12 = 7x + 22 \] \[ 7x - 3y + 10 = 0 \]
Hence, the equation of the line is: \[ 7x - 3y + 10 = 0 \]
Step 3: Condition for line to touch the circle (one point)
The line \( 7x - 3y + 10 = 0 \) is tangent to the circle. The perpendicular distance from the center \( (1, -1) \) to this line equals the radius \( r \).
Distance from \( (x_1, y_1) \) to \( ax + by + c = 0 \): \[ d = \frac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}} \]
Substitute \( a = 7, b = -3, c = 10, (x_1, y_1) = (1, -1) \): \[ d = \frac{|7(1) - 3(-1) + 10|}{\sqrt{7^2 + (-3)^2}} = \frac{|7 + 3 + 10|}{\sqrt{49 + 9}} = \frac{20}{\sqrt{58}} \] \[ r = \frac{20}{\sqrt{58}} \]
Step 4: Equation of the circle
\[ (x - 1)^2 + (y + 1)^2 = r^2 = \frac{400}{58} = \frac{200}{29} \] \] \[ (x - 1)^2 + (y + 1)^2 = \frac{200}{29} \]
Step 5: Find the point of tangency (P)
Equation of circle: \[ (x - 1)^2 + (y + 1)^2 = \frac{200}{29} \] Equation of tangent line: \[ 7x - 3y + 10 = 0 \] \[ \Rightarrow y = \frac{7x + 10}{3} \]
Substitute in circle equation: \[ (x - 1)^2 + \left(\frac{7x + 10}{3} + 1\right)^2 = \frac{200}{29} \] \[ (x - 1)^2 + \left(\frac{7x + 13}{3}\right)^2 = \frac{200}{29} \]
Simplify: \[ 9(x - 1)^2 + (7x + 13)^2 = \frac{1800}{29} \] \[ 9(x^2 - 2x + 1) + (49x^2 + 182x + 169) = \frac{1800}{29} \] \[ 58x^2 + 164x + 178 = \frac{1800}{29} \] \[ \Rightarrow 1682x^2 + 4756x + 5162 - 1800 = 0 \] \[ 1682x^2 + 4756x + 3362 = 0 \] \] \[ \text{Since line is tangent, discriminant = 0:} \] \[ (4756)^2 - 4(1682)(3362) = 0 \] \] Thus one point of contact \( P(\alpha, \beta) \).
Solve for \( x \): \[ \alpha = -\frac{b}{2a} = -\frac{4756}{2(1682)} = -\frac{4756}{3364} = -\frac{1189}{841} \]
Now \( y = \frac{7x + 10}{3} \): \[ \beta = \frac{7\left(-\frac{1189}{841}\right) + 10}{3} = \frac{-8323 + 8410}{2523} = \frac{87}{2523} = \frac{29}{841} \]
Step 6: Find \( 17\beta - \alpha \)
\[ 17\beta - \alpha = 17\left(\frac{29}{841}\right) - \left(-\frac{1189}{841}\right) \] \[ = \frac{493 + 1189}{841} = \frac{1682}{841} = 2 \]
Final Answer:
\[ \boxed{17\beta - \alpha = 2} \]
Approach Solution -2
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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