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Circle question

2019 · 12 Apr · Shift 2 · Q39
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  5. /2019 · 12 Apr · Shift 2 · Q39

Circle question

2019 · 12 Apr · Shift 2 · Q39

JEE MainMathematicsCircleMCQ+4 / −1
A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the point :
  1. A
    (1, 5)
  2. B
    ( 2, 3)
  3. C
    (3, 5)
  4. D
    (3, 10)
View written solutionFree

Correct answer: D

  1. Use the condition of touching the x-axis at (3,0)(3,0)(3,0)

A circle tangent to the x-axis at (3,0)(3,0)(3,0) has its center vertically above or below this point. So the center must be of the form (3,r)(3,r)(3,r) or (3,−r),(3,-r),(3,−r), with radius rrr.

Since the circle makes an intercept on the y-axis, let us take the center as C=(3,k),C=(3,k),C=(3,k), with radius R=∣k∣.R=|k|.R=∣k∣. Because the circle touches the x-axis at (3,0)(3,0)(3,0), the distance of the center from the x-axis equals the radius.

So the equation is (x−3)2+(y−k)2=k2.(x-3)^2+(y-k)^2=k^2.(x−3)2+(y−k)2=k2.

  1. Use the intercept on the y-axis

On the y-axis, x=0x=0x=0. Substituting into the circle: (0−3)2+(y−k)2=k2(0-3)^2+(y-k)^2=k^2(0−3)2+(y−k)2=k2 9+(y−k)2=k29+(y-k)^2=k^29+(y−k)2=k2 (y−k)2=k2−9.(y-k)^2=k^2-9.(y−k)2=k2−9.

Thus the y-axis cuts the circle at two points whose yyy-coordinates are y=k±k2−9.y=k\pm \sqrt{k^2-9}.y=k±k2−9​.

Hence the length of the intercept on the y-axis is 2k2−9.2\sqrt{k^2-9}.2k2−9​.

Given this length is 888: 2k2−9=82\sqrt{k^2-9}=82k2−9​=8 k2−9=4\sqrt{k^2-9}=4k2−9​=4 k2−9=16k^2-9=16k2−9=16 k2=25k^2=25k2=25 k=±5.k=\pm 5.k=±5.

So possible circles are:

  • center (3,5)(3,5)(3,5), radius 555
  • center (3,−5)(3,-5)(3,−5), radius 555
  1. Write the corresponding equations

For center (3,5)(3,5)(3,5): (x−3)2+(y−5)2=25.(x-3)^2+(y-5)^2=25.(x−3)2+(y−5)2=25.

For center (3,−5)(3,-5)(3,−5): (x−3)2+(y+5)2=25.(x-3)^2+(y+5)^2=25.(x−3)2+(y+5)2=25.

  1. Check the options

We test which given point lies on the circle.

Option A: (1,5)(1,5)(1,5)

Using (x−3)2+(y−5)2=25(x-3)^2+(y-5)^2=25(x−3)2+(y−5)2=25 (1−3)2+(5−5)2=4≠25.(1-3)^2+(5-5)^2=4\neq 25.(1−3)2+(5−5)2=4=25. Not on this circle. Using the other circle: (1−3)2+(5+5)2=4+100=104≠25.(1-3)^2+(5+5)^2=4+100=104\neq 25.(1−3)2+(5+5)2=4+100=104=25. So A is false.

Option B: (2,3)(2,3)(2,3)

For (x−3)2+(y−5)2=25(x-3)^2+(y-5)^2=25(x−3)2+(y−5)2=25 (2−3)2+(3−5)2=1+4=5≠25.(2-3)^2+(3-5)^2=1+4=5\neq 25.(2−3)2+(3−5)2=1+4=5=25. For (x−3)2+(y+5)2=25(x-3)^2+(y+5)^2=25(x−3)2+(y+5)2=25 (2−3)2+(3+5)2=1+64=65≠25.(2-3)^2+(3+5)^2=1+64=65\neq 25.(2−3)2+(3+5)2=1+64=65=25. So B is false.

Option C: (3,5)(3,5)(3,5)

For (x−3)2+(y−5)2=25(x-3)^2+(y-5)^2=25(x−3)2+(y−5)2=25 (3−3)2+(5−5)2=0≠25.(3-3)^2+(5-5)^2=0\neq 25.(3−3)2+(5−5)2=0=25. This is the center, not a point on the circle. For the other circle: (3−3)2+(5+5)2=100≠25.(3-3)^2+(5+5)^2=100\neq 25.(3−3)2+(5+5)2=100=25. So C is false.

Option D: (3,10)(3,10)(3,10)

For (x−3)2+(y−5)2=25(x-3)^2+(y-5)^2=25(x−3)2+(y−5)2=25 (3−3)2+(10−5)2=25.(3-3)^2+(10-5)^2=25.(3−3)2+(10−5)2=25. This satisfies the equation. So D lies on the circle.

Also for the circle centered at (3,−5)(3,-5)(3,−5), point (3,10)(3,10)(3,10) does not lie on it, but the question asks for a point through which a circle satisfying the conditions passes. Option D works.

  1. Final answer

The required point is (3,10).\boxed{(3,10)}.(3,10)​. So the correct option is D.

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