Let the circle C touch the line x−y+1=0, have the centre on the positive x -axis, and cut off a chord of length 134 along the line −3x+2y=1. Let H be the hyperbola α2x2−2αy2=1, whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C. Then 2α2+3β2 is equal to
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Step-by-Step Solution
Step 1: Find the center and radius of circle C
Let the center of the circle C be (h,0) since it lies on the positive x -axis. Let r be its radius. The distance from the center (h,0) to the tangent line x−y+1=0 is equal to the radius r. We use the formula for the distance from a point to a line. Also, the chord length formula relates the radius, the distance from the center to the chord, and half the chord length. The distance from (h,0) to the chord line −3x+2y−1=0 is d=(−3)2+22∣−3h−1∣=13∣3h+1∣. The chord length is given as 134, so half the chord length is 132. Using the Pythagorean theorem, r2=d2+(half chord length)2.
Step 2: Solve for h and r
We equate the two expressions for r2 and solve for h. Since the center is on the positive x -axis, we take the positive value of h. Once h is found, we can calculate the radius r.
Step 3: Determine hyperbola parameters
The center of circle C is (3,0). The diameter of circle C is 2r=2(22)=42. For the hyperbola α2x2−2αy2=1, we have a2=α2 and b2=2α. The length of the transverse axis is 2a, and one of the foci is (ae,0). We are given that one focus of the hyperbola is the center of C, so ae=3. We are also given that the length of the transverse axis is the diameter of C, so 2a=42.
Step 4: Calculate α and β
From the previous step, we have a=22, so a2=8. Since a2=α2, we get α2=8. Using ae=3, we find the eccentricity e. Then, we use the relationship b2=a2(e2−1) for a hyperbola. We know b2=2α, so we substitute the values to verify consistency and find β2. However, the problem statement uses β2 in the final expression, but the hyperbola equation is given as α2x2−2αy2=1. This implies b2=2α. There seems to be a typo in the problem statement, where it should be 2α instead of 2α and β2 in the final expression. Assuming the hyperbola equation is α2x2−β2y2=1, then β2=2α.
Step 5: Calculate β2 and the final expression
From the hyperbola equation, we have b2=2α. Substituting the value of α=22, we get β2=42. Now we can substitute the values of α2 and β2 into the expression 2α2+3β2.