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Circle question

2021 · 27 Jul · Shift 2 · Q36
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  5. /2021 · 27 Jul · Shift 2 · Q36

Circle question

2021 · 27 Jul · Shift 2 · Q36

JEE MainMathematicsCircleMCQ+4 / −1
Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept 656\sqrt 565​ on the x-axis. Then the radius of the circle C is equal to :
  1. A
    53\sqrt {53}53​
  2. B
    9
  3. C
    8
  4. D
    82\sqrt {82}82​
View written solutionFree

Correct answer: B

  1. Use the tangency condition with the y-axis

Since the circle touches the y-axis at (0,6)(0,6)(0,6), its center must lie horizontally from this point.

So if the center is (h,k)(h,k)(h,k) and radius is rrr, then:

  • k=6k=6k=6
  • distance from center to the y-axis =∣h∣=r=|h|=r=∣h∣=r

Hence the center is of the form

(h,6),r=∣h∣.(h,6), \quad r=|h|.(h,6),r=∣h∣.
  1. Use the intercept on the x-axis

The circle cuts the x-axis in a chord of length 656\sqrt{5}65​.

Distance of the center (h,6)(h,6)(h,6) from the x-axis is 666. For a circle of radius rrr, the chord cut by a line at distance ddd from the center has length

2r2−d2.2\sqrt{r^2-d^2}.2r2−d2​.

Here d=6d=6d=6, so

2r2−62=65.2\sqrt{r^2-6^2}=6\sqrt{5}.2r2−62​=65​.

That is,

2r2−36=652\sqrt{r^2-36}=6\sqrt{5}2r2−36​=65​ r2−36=35\sqrt{r^2-36}=3\sqrt{5}r2−36​=35​

Squaring,

r2−36=45r^2-36=45r2−36=45 r2=81r^2=81r2=81 r=9.r=9.r=9.
  1. Check with the center condition

Since r=∣h∣r=|h|r=∣h∣, we get ∣h∣=9|h|=9∣h∣=9, so possible centers are (9,6)(9,6)(9,6) or (−9,6)(-9,6)(−9,6), both valid. Thus the radius is indeed

9.\boxed{9}.9​.
  1. Option check
  • A: 53\sqrt{53}53​ ✗
  • B: 999 ✓
  • C: 888 ✗
  • D: 82\sqrt{82}82​ ✗

So the correct option is B.

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