Let the equation of the circle, which touches x -axis at the point (a,0), a>0 and cuts off an intercept of length b on y -axis be x2+y2−αx+βy+γ=0. If the circle lies below x -axis, then the ordered pair (2a,b2) is equal to
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Step-by-Step Solution
Step 1: Determine the center and radius from the tangency condition
Since the circle touches the x -axis at (a,0) and lies below the x -axis, its center must have coordinates (a, -r), where r is the radius of the circle. The y -coordinate of the center is negative because the circle is below the x -axis.
Step 2: Write the equation of the circle
Using the standard form of a circle's equation, (x−h)2+(y−k)2=r2, and substituting the center (a, -r) and radius r, we get the equation of the circle.
Step 3: Expand and compare with the given equation
Expanding the equation and simplifying, we can compare it with the given general equation x2+y2−αx+βy+γ=0. This comparison allows us to find the values of α, β, and γ in terms of a and r.
Step 4: Use the y-intercept condition
The length of the intercept cut off on the y -axis by the circle x2+y2+2gx+2fy+c=0 is given by 2f2−c. From our expanded equation, 2g=−2a, 2f=2r, and c=a2. So, f=r. The y -intercept length is 2r2−a2. However, the problem states the intercept length is b. Therefore, b=2r2−a2. This implies b2=4(r2−a2).
Step 5: Re-evaluate the y-intercept condition based on the problem statement
To find the y -intercept, we set x=0 in the circle's equation, resulting in y2+2ry+a2=0. Let the roots of this quadratic equation be y1 and y2. The length of the intercept b is ∣y1−y2∣. Using Vieta's formulas, y1+y2=−2r and y1y2=a2. Substituting these into the formula for the difference of roots, we get b=2r2−a2. Squaring both sides gives b2=4(r2−a2).
Step 6: Solve for the ordered pair
The problem asks for the ordered pair (2a,b2). We have already determined b2=4(r2−a2). Therefore, the ordered pair is (2a,4r2−4a2). The problem statement implies that b is a specific length, which means r2−a2 must be positive, so r > a. Without a specific value for r, the expression for b2 will remain in terms of r and a.