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Circle question

2018 · 16 Apr · Shift 1 · Q46
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Circle question

2018 · 16 Apr · Shift 1 · Q46

JEE MainMathematicsCircleMCQ+4 / −1
If a circle C, whose radius is 3, touches externally the circle, x2+y2+2x−4y−4=0{x^2} + {y^2} + 2x - 4y - 4 = 0x2+y2+2x−4y−4=0 at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to :
  1. A
    252\sqrt 525​
  2. B
    323\sqrt 232​
  3. C
    5\sqrt 55​
  4. D
    232\sqrt 323​
View written solutionFree

Correct answer: A

  1. Given circle and its center/radius

    The given circle is x2+y2+2x−4y−4=0.x^2+y^2+2x-4y-4=0.x2+y2+2x−4y−4=0.

    Compare with the standard form x2+y2+2gx+2fy+c=0,x^2+y^2+2gx+2fy+c=0,x2+y2+2gx+2fy+c=0, so g=1, f=−2, c=−4.g=1,\, f=-2,\, c=-4.g=1,f=−2,c=−4.

    Hence its center is (−g,−f)=(−1,2),(-g,-f)=(-1,2),(−g,−f)=(−1,2), and radius is r1=g2+f2−c=1+4+4=3.r_1=\sqrt{g^2+f^2-c}=\sqrt{1+4+4}=3.r1​=g2+f2−c​=1+4+4​=3.

  2. Use the point of contact

    The other circle CCC also has radius 333 and touches the given circle externally at (2,2)(2,2)(2,2).

    For two circles touching externally:

    • the point of contact lies on the line joining the centers,
    • the distance between centers is the sum of radii.

    Since both radii are equal to 333, the point of contact (2,2)(2,2)(2,2) is the midpoint of the two centers.

    Let the center of circle CCC be (h,k)(h,k)(h,k). Since (−1,2)(-1,2)(−1,2) and (h,k)(h,k)(h,k) have midpoint (2,2)(2,2)(2,2), (−1+h2,2+k2)=(2,2).\left(\frac{-1+h}{2},\frac{2+k}{2}\right)=(2,2).(2−1+h​,22+k​)=(2,2).

    Therefore, −1+h2=2  ⟹  h=5,\frac{-1+h}{2}=2 \implies h=5,2−1+h​=2⟹h=5, 2+k2=2  ⟹  k=2.\frac{2+k}{2}=2 \implies k=2.22+k​=2⟹k=2.

    So the center of circle CCC is (5,2),(5,2),(5,2), and its equation is (x−5)2+(y−2)2=9.(x-5)^2+(y-2)^2=9.(x−5)2+(y−2)2=9.

  3. Find intercept cut on the x-axis

    On the x-axis, y=0y=0y=0. Substitute into the circle: (x−5)2+(0−2)2=9(x-5)^2+(0-2)^2=9(x−5)2+(0−2)2=9 (x−5)2+4=9 (x-5)^2+4=9(x−5)2+4=9 (x−5)2=5 (x-5)^2=5(x−5)2=5 x=5±5.x=5\pm \sqrt{5}.x=5±5​.

    So the circle cuts the x-axis at the points (5−5,0),(5+5,0).\left(5-\sqrt{5},0\right), \quad \left(5+\sqrt{5},0\right).(5−5​,0),(5+5​,0).

    Hence the length of the intercept is (5+5)−(5−5)=25.\left(5+\sqrt{5}\right)-\left(5-\sqrt{5}\right)=2\sqrt{5}.(5+5​)−(5−5​)=25​.

  4. Check options

    252\sqrt{5}25​ matches Option A.

Final Answer: 25\boxed{2\sqrt{5}}25​​

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