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Circle question

2016 · 9 Apr · Shift 1 · Q31
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  5. /2016 · 9 Apr · Shift 1 · Q31

Circle question

2016 · 9 Apr · Shift 1 · Q31

JEE MainMathematicsCircleMCQ+4 / −1
A circle passes through (−2, 4) and touches the y-axis at (0, 2). Which one of the following equations can represent a diameter of this circle?
  1. A
    4x + 5y − 6 = 0
  2. B
    2x − 3y + 10 = 0
  3. C
    3x + 4y − 3 = 0
  4. D
    5x + 2y + 4 = 0
View written solutionFree

Correct answer: B

  1. Use the tangency condition to find the center

Since the circle touches the yyy-axis at (0,2)(0,2)(0,2), the radius at the point of contact is perpendicular to the yyy-axis.

  • The yyy-axis is the line x=0x=0x=0.
  • A line perpendicular to it is horizontal.

Therefore, the center of the circle must lie on the horizontal line through (0,2)(0,2)(0,2), so let the center be

(h,2).(h,2).(h,2).

Also, since the circle touches the yyy-axis at (0,2)(0,2)(0,2), the radius is the horizontal distance from (h,2)(h,2)(h,2) to (0,2)(0,2)(0,2), i.e.

r=∣h∣.r=|h|.r=∣h∣.
  1. Use the fact that the circle passes through (−2,4)(-2,4)(−2,4)

The distance from the center (h,2)(h,2)(h,2) to (−2,4)(-2,4)(−2,4) must also equal the radius:

(h+2)2+(2−4)2=∣h∣.\sqrt{(h+2)^2+(2-4)^2}=|h|.(h+2)2+(2−4)2​=∣h∣.

Squaring both sides,

(h+2)2+(−2)2=h2(h+2)^2+(-2)^2=h^2(h+2)2+(−2)2=h2 (h+2)2+4=h2(h+2)^2+4=h^2(h+2)2+4=h2 h2+4h+4+4=h2h^2+4h+4+4=h^2h2+4h+4+4=h2 4h+8=04h+8=04h+8=0 h=−2.h=-2.h=−2.

So the center is

(−2,2)(-2,2)(−2,2)

and the radius is

r=2.r=2.r=2.
  1. Find which given line can be a diameter

A diameter is any line passing through the center of the circle. So we check which option passes through (−2,2)(-2,2)(−2,2).

Option A: 4x+5y−6=04x+5y-6=04x+5y−6=0

4(−2)+5(2)−6=−8+10−6=−4≠04(-2)+5(2)-6=-8+10-6=-4\neq 04(−2)+5(2)−6=−8+10−6=−4=0

So A does not pass through the center.

Option B: 2x−3y+10=02x-3y+10=02x−3y+10=0

2(−2)−3(2)+10=−4−6+10=02(-2)-3(2)+10=-4-6+10=02(−2)−3(2)+10=−4−6+10=0

So B does pass through the center.

Option C: 3x+4y−3=03x+4y-3=03x+4y−3=0

3(−2)+4(2)−3=−6+8−3=−1≠03(-2)+4(2)-3=-6+8-3=-1\neq 03(−2)+4(2)−3=−6+8−3=−1=0

So C does not pass through the center.

Option D: 5x+2y+4=05x+2y+4=05x+2y+4=0

5(−2)+2(2)+4=−10+4+4=−2≠05(-2)+2(2)+4=-10+4+4=-2\neq 05(−2)+2(2)+4=−10+4+4=−2=0

So D does not pass through the center.

  1. Conclusion

The only line passing through the center, and hence capable of representing a diameter, is

2x−3y+10=0.\boxed{2x-3y+10=0}.2x−3y+10=0​.

So the correct option is B.

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