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Step-by-Step Solution
Step 1: Find the Center of the Circle
We have two diameters of the circle given by the straight lines
2x - 3y = 5 and 3x - 4y = 7 .
The intersection of these two lines is the center of the circle.
Solve them simultaneously:
From 2x - 3y = 5 , express x in terms of y :
2x = 5 + 3y
\quad\Longrightarrow\quad
x = \frac{5 + 3y}{2}.
Substitute this into 3x - 4y = 7 :
3\left(\frac{5 + 3y}{2}\right) - 4y = 7
\quad\Longrightarrow\quad
\frac{15 + 9y}{2} - 4y = 7.
Multiply through by 2 to clear the fraction:
15 + 9y - 8y = 14
\quad\Longrightarrow\quad
15 + y = 14
\quad\Longrightarrow\quad
y = -1.
Substitute y = -1 back to find x :
x = \frac{5 + 3(-1)}{2} = \frac{5 - 3}{2} = 1.
Therefore, the center of the circle is (1,\, -1) .
Step 2: Shift the Origin to the Center
To simplify the algebra, introduce new coordinates
X = x - 1, \quad Y = y + 1.
In these (X, Y) coordinates, the center of the circle at (1, -1) in the original plane becomes (0, 0) in the new plane.
Rewrite the given diameters in terms of X and Y :
1) For 2x - 3y = 5, substitute x = X + 1, \; y = Y - 1 :
2(X + 1) - 3(Y - 1) = 5
\quad\Longrightarrow\quad
2X + 2 - 3Y + 3 = 5
\quad\Longrightarrow\quad
2X - 3Y + 5 = 5
\quad\Longrightarrow\quad
2X - 3Y = 0.
2) For 3x - 4y = 7, substitute x = X + 1, \; y = Y - 1 :
3(X + 1) - 4(Y - 1) = 7
\quad\Longrightarrow\quad
3X + 3 - 4Y + 4 = 7
\quad\Longrightarrow\quad
3X - 4Y + 7 = 7
\quad\Longrightarrow\quad
3X - 4Y = 0.
Thus, in the shifted (X, Y) system, the diameters become
2X - 3Y = 0 and 3X - 4Y = 0,
both passing through the origin (0,0) .
Step 3: Equation of the Circle and Tangent Condition
Since the center is now at (0, 0) in the (X, Y) coordinates, the circle can be written as
X^2 + Y^2 = r^2,
for some radius r .
We are told that the line joining \left(-\frac{22}{7}, -4\right) and \left(-\frac{1}{7}, 3\right) touches the circle at exactly one point. First, find the equation of that line in the original (x, y) coordinates.
The slope of this line is
m = \frac{3 - (-4)}{-\frac{1}{7} - \left(-\frac{22}{7}\right)}
= \frac{7}{\frac{21}{7}}
= \frac{7}{3}.
By using point-slope form or direct derivation, one obtains the line equation
7x - 3y + 10 = 0.
Convert this into (X, Y) coordinates where x = X + 1 and y = Y - 1 :
7(X+1) - 3(Y-1) + 10 = 0
\quad\Longrightarrow\quad
7X + 7 - 3Y + 3 + 10 = 0
\quad\Longrightarrow\quad
7X - 3Y + 20 = 0.
In the (X, Y) plane, the distance from the center (0,0) to the line 7X - 3Y + 20 = 0 is
\frac{|20|}{\sqrt{7^2 + (-3)^2}}
= \frac{20}{\sqrt{49 + 9}}
= \frac{20}{\sqrt{58}}.
Since the line meets the circle at exactly one point, it must be tangent. Hence, the radius r of the circle equals this distance:
r = \frac{20}{\sqrt{58}},
\quad
r^2 = \frac{400}{58} = \frac{200}{29}.
Returning to the original (x,y) coordinates, the circle’s equation is
(x - 1)^2 + (y + 1)^2 = \frac{200}{29}.
Step 4: Find the Coordinates of the Tangency Point
We want the point P(\alpha, \beta) on the circle
(x - 1)^2 + (y + 1)^2 = \frac{200}{29}
which also satisfies the line
7x - 3y + 10 = 0.
From the line 7x - 3y + 10 = 0, express y in terms of x :
7x + 10 = 3y
\quad\Longrightarrow\quad
y = \frac{7x + 10}{3}.
Substitute this into the circle equation:
(x - 1)^2 + \left(\frac{7x + 10}{3} + 1\right)^2
= \frac{200}{29}.
Simplify inside the second term:
\frac{7x + 10}{3} + 1
= \frac{7x + 10 + 3}{3}
= \frac{7x + 13}{3}.
So the circle equation becomes
(x - 1)^2 + \left(\frac{7x + 13}{3}\right)^2
= \frac{200}{29}.
Multiply through by 9 to clear denominators:
9(x - 1)^2 + (7x + 13)^2
= 9 \times \frac{200}{29}.
Upon simplifying, one finds a single (repeated) root, because the line is tangent:
(29x + 41)^2 = 0
\quad\Longrightarrow\quad
x = -\frac{41}{29}.
Substitute back to find y :
y = \frac{7\left(-\frac{41}{29}\right) + 10}{3}
= \frac{-\frac{287}{29} + \frac{290}{29}}{3}
= \frac{\frac{3}{29}}{3}
= \frac{1}{29}.
Hence, the point of tangency P(\alpha, \beta) is
\left(-\frac{41}{29}, \frac{1}{29}\right).
Step 5: Compute 17\beta - \alpha
From P(\alpha, \beta) , we have \alpha = -\frac{41}{29} and \beta = \frac{1}{29}.
Therefore:
17\beta - \alpha
= 17 \times \frac{1}{29} - \left(-\frac{41}{29}\right)
= \frac{17}{29} + \frac{41}{29}
= \frac{58}{29}
= 2.
So, the required value of 17\beta - \alpha is 2 .
Final Answer
17\beta - \alpha = 2.
