Let C1 and C2 be two circles of radius r1 and r2 respectively (r1 r2) touching both the axes. If the two circles are orthogonal, then (r1/r2) is equal to

Question

Let C1 and C2 be two circles of radius r1 and r2 respectively (r1 > r2) touching both the axes. If the two circles are orthogonal, then (r1/r2) is equal to (A) 2 (B) 2+√3 (C) 3+√2 (D) 3

Tardigrade Admin · Accepted Answer

Let the equation of a circle touching both the axes be (x - r)2+(y - r)2=r2 x2+y2-2rx-2ry+r2=0 If the two given circles are orthogonal, then 2(- r1)(- r2)+2(- r1)(- r2)=r12+r22 ⇒ r12-4r1r2+r22=0 ⇒ (r1/r2)=2+√3

Q. Let $C_{1}$ and $C_{2}$ be two circles of radius $r_{1}$ and $r_{2}$ respectively $(r_{1} > r_{2})$ touching both the axes. If the two circles are orthogonal, then $\frac{r _{1}}{r _{2}}$ is equal to

$2$

$2 + 3$

$3 + 2$

$3$

Q. Let C1​ and C2​ be two circles of radius r1​ and r2​ respectively (r1​>r2​) touching both the axes. If the two circles are orthogonal, then r2​r1​​ is equal to

2

2+3​

3+2​

3

Let the equation of a circle touching both the axes be (x−r)2+(y−r)2=r2 x2+y2−2rx−2ry+r2=0 If the two given circles are orthogonal, then 2(−r1​)(−r2​)+2(−r1​)(−r2​)=r12​+r22​ ⇒r12​−4r1​r2​+r22​=0 ⇒r2​r1​​=2+3​

Q. Let $C_{1}$ and $C_{2}$ be two circles of radius $r_{1}$ and $r_{2}$ respectively $(r_{1} > r_{2})$ touching both the axes. If the two circles are orthogonal, then $\frac{r _{1}}{r _{2}}$ is equal to

$2$

$2 + 3$

$3 + 2$

$3$