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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Circle

The absolute difference between the squares of the radii of the two...

The absolute difference between the squares of the radii of the two circles passing through the point (−9,4)(-9,4) and touching the lines x+y=3x+y=3 and x−y=3x-y=3, is equal to ______.The absolute difference between the squares of the radii of the two circles passing through the point (-9,4) and

Correct answer

0

Step-by-step solution

The absolute difference between the squares of the radii of the two circles passing through the point (-9,4) and
Centre (a,0)(\mathrm{a}, 0)
r=∣a−0−32∣\mathrm{r}=\left|\frac{\mathrm{a}-0-3}{\sqrt{2}}\right|
circle⁡(x−a)2+y2=(a−32)2\operatorname{circle}(x-a)^2+y^2=\left(\frac{a-3}{\sqrt{2}}\right)^2
passes through (−9,4)(-9,4)
2(a2+18a+81+16)=(a2−6a+9)2\left(a^2+18 a+81+16\right)=\left(a^2-6 a+9\right)
a2+42a+185=0\mathrm{a}^2+42 \mathrm{a}+185=0
(a+37)(a+5)=0(a+37)(a+5)=0
⇒a=−37,−5\Rightarrow a=-37,-5
r1=∣−37−32∣=202r_1=\left|\frac{-37-3}{\sqrt{2}}\right|=20 \sqrt{2}
r2=∣−5−32∣=42\mathrm{r}_2=\left|\frac{-5-3}{\sqrt{2}}\right|=4 \sqrt{2}
\left|r_1^2-r_2^2\right|=|800-32|=768
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