Q.

Let r be the radius of the circle, which touches x-axis at point (a, 0), a < 0 and the parabola y2=9x 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=r2

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

         (4–a)2+(6–r)2=r2

⇒16+a2–8a+36+r2–12r=r2

⇒ a2–12r–8a+52=0          ... (i)

Equation of tangent to the barabola y2=9x at the point A(4, 6) is given by yy1=2a(x+x1)

⇒ 6y=2×94(x+4) ⇒ 6y=9x+362

⇒ 3x–4y+12=0

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

∴  |3a–4r+129+16|=r

⇒ 3a–4r+12=±5r

⇒ 3a+12=9r or 3a+12=-r

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

a2–12(a+43)–8a+52=0

⇒ a2–4a–16–8a+52=0 ⇒ a2–12a+36=0

⇒ (a–6)2=0 ⇒ a=6 but a<0 ⇒ a≠6

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

         a2–8a+12(3a+12)+52=0

⇒ a2+28a+196=0 ⇒ (a+14)2=0

⇒ a=–14 ⇒ r=30.