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Circle question

2010 · Shift 0 · Q42
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  5. /2010 · Shift 0 · Q42

Circle question

2010 · Shift 0 · Q42

JEE MainMathematicsCircleMCQ+4 / −1
The circle x2+y2=4x+8y+5{x^2} + {y^2} = 4x + 8y + 5x2+y2=4x+8y+5 intersects the line 3x−4y=m3x - 4y = m3x−4y=m at two distinct points if :
  1. A
    −35<m<15- 35 \lt m \lt 15−35<m<15
  2. B
    15<m<6515 \lt m \lt 6515<m<65
  3. C
    35<m<8535 \lt m \lt 8535<m<85
  4. D
    −85<m<−35- 85 \lt m \lt -35−85<m<−35
View written solutionFree

Correct answer: A

  1. Write the circle in standard form

Given circle: x2+y2=4x+8y+5x^2+y^2=4x+8y+5x2+y2=4x+8y+5

Bring all terms to one side: x2−4x+y2−8y−5=0x^2-4x+y^2-8y-5=0x2−4x+y2−8y−5=0

Complete squares: x2−4x=(x−2)2−4x^2-4x=(x-2)^2-4x2−4x=(x−2)2−4 y2−8y=(y−4)2−16y^2-8y=(y-4)^2-16y2−8y=(y−4)2−16

So, (x−2)2−4+(y−4)2−16−5=0(x-2)^2-4+(y-4)^2-16-5=0(x−2)2−4+(y−4)2−16−5=0 (x−2)2+(y−4)2=25(x-2)^2+(y-4)^2=25(x−2)2+(y−4)2=25

Hence, the circle has:

  • Centre C(2,4)C(2,4)C(2,4)
  • Radius r=5r=5r=5
  1. Condition for a line to intersect a circle at two distinct points

The line is: 3x−4y=m3x-4y=m3x−4y=m

For the line to cut the circle at two distinct points, the perpendicular distance from the centre to the line must be less than the radius.

So we need: distance from (2,4) to 3x−4y−m=0<5\text{distance from }(2,4)\text{ to }3x-4y-m=0<5distance from (2,4) to 3x−4y−m=0<5

Distance formula: d=∣3(2)−4(4)−m∣32+(−4)2d=\frac{|3(2)-4(4)-m|}{\sqrt{3^2+(-4)^2}}d=32+(−4)2​∣3(2)−4(4)−m∣​

Compute: d=\frac{|6-16-m|}{5}= rac{|-10-m|}{5}= rac{|m+10|}{5}

For two distinct intersections: ∣m+10∣5<5\frac{|m+10|}{5}<55∣m+10∣​<5

Multiply by 555: ∣m+10∣<25|m+10|<25∣m+10∣<25

This gives: −25<m+10<25-25<m+10<25−25<m+10<25 −35<m<15-35<m<15−35<m<15

  1. Match with the options

The required range is: −35<m<15-35<m<15−35<m<15

This is Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They match.

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