Let $r$ be the radius of the circle, which touches the x-axis at point $(a, 0), \; a < 0$ and the parabola $y^2 = 9x$ at the point $(4, 6)$. Then $r$ is equal to _____

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 $y^2 = 9x$ 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 $(x - a)^2 + (y - r)^2 = r^2$.

Step 2: Use the point of tangency on the parabola

The circle touches the parabola $y^2 = 9x$ 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 $y^2 = 9x$ at $(4, 6)$, we differentiate implicitly with respect to $x$. This gives $2y \frac{dy}{dx} = 9$, so $\frac{dy}{dx} = \frac{9}{2y}$. Substituting $y=6$, we get the slope of the tangent to the parabola, $m_p = \frac{3}{4}$.

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, $m_r$, 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, $m_r = -\frac{1}{m_p}$. 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, $a = \frac{34 - 3r}{4}$. Substitute this expression for $a$ into the circle equation $(4 - a)^2 + (6 - r)^2 = r^2$. This yields a quadratic equation in $r$: $25r^2 - 250r + 625 = 0$, which simplifies to $(r - 5)^2 = 0$. Thus, $r = 5$. We can then find $a = \frac{34 - 3(5)}{4} = \frac{19}{4}$. However, the problem states $a < 0$. Let's recheck the center. If the center is $(a, -r)$, then the equation is $(x-a)^2 + (y+r)^2 = r^2$. The point $(4{,}6)$ is on the circle, so $(4-a)^2 + (6+r)^2 = r^2$. The slope of the radius is $\frac{6 - (-r)}{4 - a} = \frac{6+r}{4-a}$. This must be equal to $-\frac{1}{m_p} = -\frac{4}{3}$. So $3(6+r) = -4(4-a)$, which means $18+3r = -16+4a$, or $4a = 34+3r$. Substituting this into the circle equation: $(4 - \frac{34+3r}{4})^2 + (6+r)^2 = r^2$. This simplifies to $(\frac{16 - 34 - 3r}{4})^2 + (6+r)^2 = r^2 \implies (\frac{-18 - 3r}{4})^2 + (6+r)^2 = r^2 \implies \frac{9(6+r)^2}{16} + (6+r)^2 = r^2$. Let $X = 6+r$. Then $\frac{9X^2}{16} + X^2 = r^2 \implies \frac{25X^2}{16} = r^2 \implies \frac{25(6+r)^2}{16} = r^2$. Taking the square root of both sides: $\frac{5(6+r)}{4} = \pm r$. Case 1: $\frac{5(6+r)}{4} = r \implies 30 + 5r = 4r \implies r = -30$. Radius cannot be negative. Case 2: $\frac{5(6+r)}{4} = -r \implies 30 + 5r = -4r \implies 9r = -30 \implies r = -\frac{30}{9} = -\frac{10}{3}$. 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 $m_p = 3/4$. The slope of the radius from $(a,r)$ to $(4{,}6)$ is $m_r = \frac{6-r}{4-a}$. Since the radius is perpendicular to the tangent, $m_r = -1/m_p = -4/3$. So, $\frac{6-r}{4-a} = -\frac{4}{3}$. $3(6-r) = -4(4-a)$ $18 - 3r = -16 + 4a$ $4a = 34 - 3r \implies a = \frac{34 - 3r}{4}$. Now substitute this into the circle equation $(4-a)^2 + (6-r)^2 = r^2$. $(4 - \frac{34 - 3r}{4})^2 + (6-r)^2 = r^2$ $(\frac{16 - 34 + 3r}{4})^2 + (6-r)^2 = r^2$ $(\frac{3r - 18}{4})^2 + (6-r)^2 = r^2$ $\frac{9(r - 6)^2}{16} + (r - 6)^2 = r^2$ $(r - 6)^2 (\frac{9}{16} + 1) = r^2$ $(r - 6)^2 (\frac{25}{16}) = r^2$ Take the square root of both sides: $\frac{5}{4} (r - 6) = \pm r$ Case 1: $\frac{5}{4} (r - 6) = r$ $5r - 30 = 4r$ $r = 30$ If $r = 30$, then $a = \frac{34 - 3(30)}{4} = \frac{34 - 90}{4} = \frac{-56}{4} = -14$. This satisfies $a < 0$. So $r=30$ is a valid solution. Case 2: $\frac{5}{4} (r - 6) = -r$ $5r - 30 = -4r$ $9r = 30$ $r = \frac{30}{9} = \frac{10}{3}$ If $r = \frac{10}{3}$, then $a = \frac{34 - 3(\frac{10}{3})}{4} = \frac{34 - 10}{4} = \frac{24}{4} = 6$. This does not satisfy $a < 0$. So $r = \frac{10}{3}$ is not a valid solution. Therefore, $r = 30$.

More from this chapter

• Find the equation of the circle passing through points (1,2), (3,4), and (5,2).
• Determine the equation of the circle with radius 4 and Centre (-2, 3).
• Find the equation of the hyperbola where foci are (0, +-6) and the length of the latus rectum is 36.
• Find the equation of the hyperbola where foci are (0, ± 12) and the length of the latus rectum is 36
• Find the equation of the circle with centre (-3,2) and radius 10.
• Find the equation of the circle with centre (-3,2) and radius 9.
• Find the equation of the circle with centre (-3,2) and radius 8.
• Find the equation of the circle with centre (-3,2) and radius 6.

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