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Circle question

2019 · Shift 1 · Q21
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  5. /2019 · Shift 1 · Q21

Circle question

2019 · Shift 1 · Q21

JEE AdvancedMathematicsCircleMCQ+3 / −1
A line y = mx + 1 intersects the circle (x−3)2+(y+2)2{(x - 3)^2} + {(y + 2)^2}(x−3)2+(y+2)2= 25 at the points P and Q. If the midpoint of the line segment PQ has x-coordinate −35- {3 \over 5}−53​, then which one of the following options is correct?
  1. A
    6 ≤\le≤ m < 8
  2. B
    −-− 3 ≤\le≤ m <−-− 1
  3. C
    4 ≤\le≤ m < 6
  4. D
    2 ≤\le≤ m < 4
View written solutionFree

Correct answer: NONE OF THE GIVEN OPTIONS IS CORRECT., IF FORCED TO REPORT THE TRUE SOLUTION SET, $M=\DISPLAYSTYLE \FRAC{15\PM\SQRT{161}}{2}$.

  1. Write the circle and line in standard form

The circle is

(x−3)2+(y+2)2=25(x-3)^2+(y+2)^2=25(x−3)2+(y+2)2=25

so its center is

C=(3,−2).C=(3,-2).C=(3,−2).

The line is

y=mx+1.y=mx+1.y=mx+1.
  1. Use the fact about midpoint of a chord

If a line cuts a circle at points PPP and QQQ, then the midpoint of chord PQPQPQ is the foot of the perpendicular from the center of the circle to the line.

So, if MMM is the midpoint of PQPQPQ, then MMM is the projection of C=(3,−2)C=(3,-2)C=(3,−2) onto the line

y=mx+1.y=mx+1.y=mx+1.

We are given that the xxx-coordinate of MMM is

- rac35.
  1. Equation of the perpendicular through the center

The given line has slope mmm, so the perpendicular slope is −1m-\frac1m−m1​.

Thus the perpendicular from (3,−2)(3,-2)(3,−2) is

y+2=−1m(x−3).y+2=-\frac1m(x-3).y+2=−m1​(x−3).

The midpoint MMM lies on both lines:

y=mx+1y=mx+1y=mx+1

and

y+2=−1m(x−3).y+2=-\frac1m(x-3).y+2=−m1​(x−3).

Substitute y=mx+1y=mx+1y=mx+1 into the perpendicular equation:

mx+1+2=−1m(x−3)mx+1+2=-\frac1m(x-3)mx+1+2=−m1​(x−3) mx+3=−x−3m.mx+3=-\frac{x-3}{m}.mx+3=−mx−3​.

Multiply by mmm:

m2x+3m=−x+3.m^2x+3m=-x+3.m2x+3m=−x+3.

So,

(m2+1)x=3−3m.(m^2+1)x=3-3m.(m2+1)x=3−3m.

Hence the xxx-coordinate of the midpoint is

xM=3−3mm2+1=3(1−m)m2+1.x_M=\frac{3-3m}{m^2+1}=\frac{3(1-m)}{m^2+1}.xM​=m2+13−3m​=m2+13(1−m)​.
  1. Use the given value of the midpoint's x-coordinate

Given

3(1−m)m2+1=−35.\frac{3(1-m)}{m^2+1}=-\frac35.m2+13(1−m)​=−53​.

Cross-multiply:

15(1−m)=−(m2+1).15(1-m)=-(m^2+1).15(1−m)=−(m2+1). 15−15m=−m2−1.15-15m=-m^2-1.15−15m=−m2−1.

Bring all terms to one side:

m2−15m+16=0.m^2-15m+16=0.m2−15m+16=0.

Solve:

m=15±225−642=15±1612.m=\frac{15\pm\sqrt{225-64}}{2}=\frac{15\pm\sqrt{161}}{2}.m=215±225−64​​=215±161​​.

Now,

161≈12.69.\sqrt{161}\approx 12.69.161​≈12.69.

So,

m≈15+12.692=13.845m\approx \frac{15+12.69}{2}=13.845m≈215+12.69​=13.845

and

m≈15−12.692=1.155.m\approx \frac{15-12.69}{2}=1.155.m≈215−12.69​=1.155.
  1. Choose the correct interval

Among the given options, only

1.155∈[2,4)1.155 \in [2,4)1.155∈[2,4)

is false, and

13.84513.84513.845

fits none of the options.

So this suggests we should re-check carefully.

  1. Alternative direct method using chord midpoint formula

Substitute y=mx+1y=mx+1y=mx+1 into the circle:

(x−3)2+(mx+3)2=25.(x-3)^2+(mx+3)^2=25.(x−3)2+(mx+3)2=25.

Expand:

x2−6x+9+m2x2+6mx+9=25x^2-6x+9+m^2x^2+6mx+9=25x2−6x+9+m2x2+6mx+9=25 (1+m2)x2+6(m−1)x−7=0.(1+m^2)x^2+6(m-1)x-7=0.(1+m2)x2+6(m−1)x−7=0.

If the intersection points have xxx-coordinates x1,x2x_1,x_2x1​,x2​, then midpoint MMM has

xM=x1+x22.x_M=\frac{x_1+x_2}{2}.xM​=2x1​+x2​​.

Using sum of roots,

x1+x2=−6(m−1)1+m2=6(1−m)1+m2.x_1+x_2=-\frac{6(m-1)}{1+m^2}=\frac{6(1-m)}{1+m^2}.x1​+x2​=−1+m26(m−1)​=1+m26(1−m)​.

Therefore

xM=3(1−m)1+m2.x_M=\frac{3(1-m)}{1+m^2}.xM​=1+m23(1−m)​.

Again,

3(1−m)1+m2=−35\frac{3(1-m)}{1+m^2}=-\frac351+m23(1−m)​=−53​

which gives

m2−15m+16=0.m^2-15m+16=0.m2−15m+16=0.

So the solutions remain

m=15±1612.m=\frac{15\pm\sqrt{161}}{2}.m=215±161​​.
  1. Compare with options

The smaller root is about 1.1551.1551.155, which lies in none of A, B, C, D. The larger root is about 13.84513.84513.845, also lies in none of A, B, C, D.

Hence the options do not match the mathematically derived result.

Therefore, the stored correct answer DDD is not consistent with the given question data.

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