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Circle question

2022 · 29 Jul · Shift 2 · Q41
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  5. /2022 · 29 Jul · Shift 2 · Q41

Circle question

2022 · 29 Jul · Shift 2 · Q41

JEE MainMathematicsCircleNumerical+4 / −1
Let ABA BAB be a chord of length 12 of the circle (x−2)2+(y+1)2=1694(x-2)^{2}+(y+1)^{2}=\frac{169}{4}(x−2)2+(y+1)2=4169​. If tangents drawn to the circle at points AAA and BBB intersect at the point PPP, then five times the distance of point PPP from chord ABA BAB is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 72

  1. Identify the circle and its radius

The circle is

(x−2)2+(y+1)2=1694.(x-2)^2+(y+1)^2=\frac{169}{4}.(x−2)2+(y+1)2=4169​.

So its center is

C=(2,−1),C=(2,-1),C=(2,−1),

and radius is

r=1694=132.r=\sqrt{\frac{169}{4}}=\frac{13}{2}.r=4169​​=213​.
  1. Use the chord-length formula to find the distance of the chord from the center

If a chord of a circle of radius rrr has length LLL, and its distance from the center is ddd, then

L=2r2−d2.L=2\sqrt{r^2-d^2}.L=2r2−d2​.

Given L=12L=12L=12, we get

12=2r2−d2  ⟹  6=r2−d2.12=2\sqrt{r^2-d^2} \implies 6=\sqrt{r^2-d^2}.12=2r2−d2​⟹6=r2−d2​.

Squaring,

r2−d2=36.r^2-d^2=36.r2−d2=36.

Now,

r2=1694,r^2=\frac{169}{4},r2=4169​,

so

d2=1694−36=169−1444=254.d^2=\frac{169}{4}-36=\frac{169-144}{4}=\frac{25}{4}.d2=4169​−36=4169−144​=425​.

Hence

d=52.d=\frac{5}{2}.d=25​.

So the perpendicular distance from the center CCC to chord ABABAB is 52\frac{5}{2}25​.

  1. Use the pole-polar relation / chord of contact property

Let PPP be the intersection point of tangents at AAA and BBB. Then ABABAB is the chord of contact of tangents drawn from PPP.

For a circle with center CCC and radius rrr, if the distance from the center to the chord of contact is ddd, and the distance from the center to the external point PPP is CPCPCP, then

d=r2CP.d=\frac{r^2}{CP}.d=CPr2​.

Thus,

CP=r2d.CP=\frac{r^2}{d}.CP=dr2​.

Substitute the values:

CP=169/45/2=1694⋅25=16910.CP=\frac{169/4}{5/2}=\frac{169}{4}\cdot\frac{2}{5}=\frac{169}{10}.CP=5/2169/4​=4169​⋅52​=10169​.
  1. Find the distance of point PPP from chord ABABAB

The center CCC, the midpoint of chord ABABAB, and point PPP lie on the same perpendicular to the chord. Since the chord lies between CCC and PPP, the required distance is

CP−d=16910−52.CP-d=\frac{169}{10}-\frac{5}{2}.CP−d=10169​−25​.

Now,

52=2510,\frac{5}{2}=\frac{25}{10},25​=1025​,

so

distance from P to chord AB=16910−2510=14410=725.\text{distance from }P\text{ to chord }AB=\frac{169}{10}-\frac{25}{10}=\frac{144}{10}=\frac{72}{5}.distance from P to chord AB=10169​−1025​=10144​=572​.
  1. Multiply by 5

Therefore,

5×725=72.5\times \frac{72}{5}=72.5×572​=72.

So the required integer is

72.\boxed{72}.72​.
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