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Circle question

2019 · Shift 1 · Q32
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  5. /2019 · Shift 1 · Q32

Circle question

2019 · Shift 1 · Q32

JEE AdvancedMathematicsCircleNumerical+3 / −1
Let the point B be the reflection of the point A(2, 3) with respect to the line 8x−6y−23=08x - 6y - 23 = 08x−6y−23=0. Let ΓA\Gamma_{A}ΓA​ and ΓB\Gamma_{B}ΓB​ be circles of radii 2 and 1 with centres A and B respectively. Let T be a common tangent to the circles ΓA\Gamma_{A}ΓA​ and ΓB\Gamma_{B}ΓB​ such that both the circles are on the same side of T. If C is the point of intersection of T and the line passing through A and B, then the length of the line segment AC is .................
Numerical answer
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Correct answer: 10

  1. Find the reflection point BBB of A(2,3)A(2,3)A(2,3) in the line 8x−6y−23=08x-6y-23=08x−6y−23=0.

For a point (x0,y0)(x_0,y_0)(x0​,y0​) reflected in the line ax+by+c=0ax+by+c=0ax+by+c=0, the reflected point is

(x0−2a(ax0+by0+c)a2+b2,  y0−2b(ax0+by0+c)a2+b2).\left(x_0-\frac{2a(ax_0+by_0+c)}{a^2+b^2},\; y_0-\frac{2b(ax_0+by_0+c)}{a^2+b^2}\right).(x0​−a2+b22a(ax0​+by0​+c)​,y0​−a2+b22b(ax0​+by0​+c)​).

Here, a=8,b=−6,c=−23,(x0,y0)=(2,3).a=8,\quad b=-6,\quad c=-23,\quad (x_0,y_0)=(2,3).a=8,b=−6,c=−23,(x0​,y0​)=(2,3).

First compute:

ax0+by0+c=8(2)−6(3)−23=16−18−23=−25.ax_0+by_0+c=8(2)-6(3)-23=16-18-23=-25.ax0​+by0​+c=8(2)−6(3)−23=16−18−23=−25.

Also,

a2+b2=82+(−6)2=64+36=100.a^2+b^2=8^2+(-6)^2=64+36=100.a2+b2=82+(−6)2=64+36=100.

So,

B=(2−2⋅8⋅(−25)100,  3−2⋅(−6)⋅(−25)100).B=\left(2-\frac{2\cdot 8\cdot(-25)}{100},\; 3-\frac{2\cdot(-6)\cdot(-25)}{100}\right).B=(2−1002⋅8⋅(−25)​,3−1002⋅(−6)⋅(−25)​).

Thus,

B=(2+4,  3−3)=(6,0).B=\left(2+4,\;3-3\right)=(6,0).B=(2+4,3−3)=(6,0).
  1. Find the distance ABABAB.
AB=(6−2)2+(0−3)2=16+9=5.AB=\sqrt{(6-2)^2+(0-3)^2}=\sqrt{16+9}=5.AB=(6−2)2+(0−3)2​=16+9​=5.

So the centers are 555 units apart.

  1. Use the geometry of the common tangent.

The circles have radii:

rA=2,rB=1.r_A=2,\qquad r_B=1.rA​=2,rB​=1.

Since both circles are on the same side of the tangent TTT, this is a direct common tangent.

Let CCC be the intersection of the tangent TTT with the line through AAA and BBB. Then CCC is the external center of similarity of the two circles. Hence,

ACBC=rArB=21.\frac{AC}{BC}=\frac{r_A}{r_B}=\frac{2}{1}.BCAC​=rB​rA​​=12​.

Also, since CCC lies outside the segment ABABAB on the side of the larger circle,

AC−BC=AB=5.AC-BC=AB=5.AC−BC=AB=5.

Let

BC=x  ⟹  AC=2x.BC=x \implies AC=2x.BC=x⟹AC=2x.

Then,

2x−x=5  ⟹  x=5.2x-x=5 \implies x=5.2x−x=5⟹x=5.

Therefore,

AC=2x=10.AC=2x=10.AC=2x=10.
  1. Final answer
10\boxed{10}10​

This matches the stored correct answer.

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