Let r be the radius of the circle, which touches x-axis at point (a, 0), a < 0 and the parabola at the point (4, 6). Then r is equal to __________. [2025]

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Solution

Q.
Let r be the radius of the circle, which touches x-axis at point (a, 0), a < 0 and the parabola $y^{2} = 9 x$ at the point (4, 6). Then r is equal to __________. [2025]

Ans.
(30)

Equation of circle is given by

${(x - a)}^{2} + {(y - r)}^{2} = r^{2}$

Now, circle passes through the point A(4, 6). So, we have

${(4 - a)}^{2} + {(6 - r)}^{2} = r^{2}$

$\Rightarrow 16 + a^{2} - 8 a + 36 + r^{2} - 12 r = r^{2}$

$\Rightarrow a^{2} - 12 r - 8 a + 52 = 0$ ... (i)

Equation of tangent to the barabola $y^{2} = 9 x$ at the point A(4, 6) is given by $y y_{1} = 2 a (x + x_{1})$

$\Rightarrow 6 y = \frac{2 \times 9}{4} (x + 4) \Rightarrow 6 y = \frac{9 x + 36}{2}$

$\Rightarrow 3 x - 4 y + 12 = 0$

Distance of this line from centre of circle is equal to radius of circle.

$∴$ $| \frac{3 a - 4 r + 12}{\sqrt{9 + 16}} |$ $= r$

$\Rightarrow 3 a - 4 r + 12 = \pm 5 r$

$\Rightarrow 3 a + 12 = 9 r or 3 a + 12 = - r$

If 3a + 12 = 9r i.e., a + 4 = 3r, then by using equation (i), we get

$a^{2} - 12 (\frac{a + 4}{3}) - 8 a + 52 = 0$

$\Rightarrow a^{2} - 4 a - 16 - 8 a + 52 = 0 \Rightarrow a^{2} - 12 a + 36 = 0$

$\Rightarrow {(a - 6)}^{2} = 0 \Rightarrow a = 6 but a < 0 \Rightarrow a \neq 6$

Now, if 3a + 12 = – r, so by using equation (i), we get

$a^{2} - 8 a + 12 (3 a + 12) + 52 = 0$

$\Rightarrow a^{2} + 28 a + 196 = 0 \Rightarrow {(a + 14)}^{2} = 0$

$\Rightarrow a = - 14 \Rightarrow r = 30$ .

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