Let r be the radius of the circle, which touches the x-axis at point (a,0),a<0 and the parabola y2=9x at the point (4,6). Then r is equal to _____
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Step-by-Step Solution
Step 1: Determine the equation of the circle
Since the circle touches the x -axis at (a,0), its center must be (a, r) or (a, -r). Given that the parabola y2=9x is in the first quadrant at (4,6), the circle must be above the x -axis, so its center is (a, r) and its radius is r. The equation of such a circle is (x−a)2+(y−r)2=r2.
Step 2: Use the point of tangency on the parabola
The circle touches the parabola y2=9x at the point (4,6). This means the point (4,6) lies on the circle. Substitute x=4 and y=6 into the circle's equation.
Step 3: Find the slope of the tangent to the parabola
To find the slope of the tangent to the parabola y2=9x at (4,6), we differentiate implicitly with respect to x. This gives 2ydxdy=9, so dxdy=2y9. Substituting y=6, we get the slope of the tangent to the parabola, mp=43.
Step 4: Find the slope of the radius to the point of tangency
The radius from the center (a, r) to the point of tangency (4,6) is perpendicular to the tangent line at (4,6). The slope of this radius, mr, is given by the change in y divided by the change in x.
Step 5: Use the perpendicularity condition
Since the radius is perpendicular to the tangent, the product of their slopes is −1. Therefore, mr=−mp1. This gives us a relationship between a and r.
Step 6: Solve the system of equations for r
From the perpendicularity condition, we have 3(6−r)=−4(4−a), which simplifies to 18−3r=−16+4a, or 4a=34−3r. So, a=434−3r. Substitute this expression for a into the circle equation (4−a)2+(6−r)2=r2. This yields a quadratic equation in r: 25r2−250r+625=0, which simplifies to (r−5)2=0. Thus, r=5. We can then find a=434−3(5)=419. However, the problem states a<0. Let's recheck the center. If the center is (a, -r), then the equation is (x−a)2+(y+r)2=r2. The point (4,6) is on the circle, so (4−a)2+(6+r)2=r2. The slope of the radius is 4−a6−(−r)=4−a6+r. This must be equal to −mp1=−34. So 3(6+r)=−4(4−a), which means 18+3r=−16+4a, or 4a=34+3r. Substituting this into the circle equation: (4−434+3r)2+(6+r)2=r2. This simplifies to (416−34−3r)2+(6+r)2=r2⟹(4−18−3r)2+(6+r)2=r2⟹169(6+r)2+(6+r)2=r2. Let X=6+r. Then 169X2+X2=r2⟹1625X2=r2⟹1625(6+r)2=r2. Taking the square root of both sides: 45(6+r)=±r. Case 1: 45(6+r)=r⟹30+5r=4r⟹r=−30. Radius cannot be negative. Case 2: 45(6+r)=−r⟹30+5r=−4r⟹9r=−30⟹r=−930=−310. Radius cannot be negative.
Let's re-evaluate the center. The circle touches the x-axis at (a,0). The center is (a, r) or (a, -r). Since the point (4,6) is on the circle, and r must be positive, the center must be (a, r) if the circle is above the x-axis, or (a, -r) if it's below. Given (4,6) is in the first quadrant, the circle must be above the x-axis, so the center is (a, r).
Let's re-examine the perpendicularity condition.
The slope of the tangent to the parabola at (4,6) is mp=3/4.
The slope of the radius from (a,r) to (4,6) is mr=4−a6−r.
Since the radius is perpendicular to the tangent, mr=−1/mp=−4/3.
So, 4−a6−r=−34.
3(6−r)=−4(4−a) 18−3r=−16+4a 4a=34−3r⟹a=434−3r.
Now substitute this into the circle equation (4−a)2+(6−r)2=r2.
(4−434−3r)2+(6−r)2=r2 (416−34+3r)2+(6−r)2=r2 (43r−18)2+(6−r)2=r2 169(r−6)2+(r−6)2=r2 (r−6)2(169+1)=r2 (r−6)2(1625)=r2
Take the square root of both sides:
45(r−6)=±r
Case 1: 45(r−6)=r 5r−30=4r r=30
If r=30, then a=434−3(30)=434−90=4−56=−14.
This satisfies a<0. So r=30 is a valid solution.
Case 2: 45(r−6)=−r 5r−30=−4r 9r=30 r=930=310
If r=310, then a=434−3(310)=434−10=424=6.
This does not satisfy a<0. So r=310 is not a valid solution.
Therefore, r=30.