Question bank/JEE Main/Mathematics

JEE MainMathematicsEasyCircles

Line L touches circles C_1, C and C_2 having radii r_1, r,
r_2 respectively. These circle pair wise touch each-other externally.
If r_1 = 36 and r_2 = 900 then find the value of
nwljs o`Ùk dks cká Li'kZ djrh gSA ¼ fp=kuqlkj½;fn r_1 = 36
rFkk r_2 = 900 gks rks [IMAGE]

Show answer and solution
AB, BC and AC are direct common tangents.
AB, BC rFkk AC,sa gS
Length of DCT =d2−(r1−r2)2= \sqrt{d^{2} - \left( r_{1} - r_{2} \right)^{2}}
=d2−(r1−r2)2= \sqrt{d^{2} - \left( r_{1} - r_{2} \right)^{2}}
where d is distance between centres
tgka d ds nksauksa dsUnzksa dh nqjh gS
Also if circles touch each other externally d= r_1+ r_2
djs rks d= r_1+ r_2
∴\thereforeIn this case DCT
=(r1+r2)2−(r1−r2)2=2r1r2= \sqrt{\left( r_{1} + r_{2} \right)^{2} - \left( r_{1} - r_{2} \right)^{2}} = 2\sqrt{r_{1}r_{2}}
_∴bl fLFkfr js
=(r1+r2)2−(r1−r2)2=2r1r2= \sqrt{\left( r_{1} + r_{2} \right)^{2} - \left( r_{1} - r_{2} \right)^{2}} = 2\sqrt{r_{1}r_{2}}
AC=2r1r2AC = 2\sqrt{r_{1}r_{2}}
⇒AB+BC=AC\Rightarrow AB + BC = AC
⇒2r(r1+r2)=2r1r2\Rightarrow 2\sqrt{r}\left( \sqrt{r_{1}} + \sqrt{r_{2}} \right) = 2\sqrt{r_{1}r_{2}}
∴r=r1r2r1+r2=36×90036+900=6×306+30=5\therefore\sqrt{r} = \frac{\sqrt{r_{1}r_{2}}}{\sqrt{r_{1}} + \sqrt{r_{2}}} = \frac{\sqrt{36 \times 900}}{\sqrt{36} + \sqrt{900}} = \frac{6 \times 30}{6 + 30} = 5

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans