Two circles of radii r1 and r2 are both touching the coordinate axes and intersecting each other orthogonally. The value of \frac{r_{1}}{r

Engineering

Mathematics

Orthogonality of two Circles

Question

Two circles of radii r₁ and r₂ are both touching the coordinate axes and intersecting each other orthogonally. The value of $\frac{r_{1}}{r_{2}}$ (where r₁ > r₂) equals

2 + $\sqrt{3}$

$\sqrt{3}$ + 1

2 – $\sqrt{3}$

2 + $\sqrt{5}$

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Circle is (x – r)² + (y – r)² = r²
⇒ x² + y² – 2xr – 2yr + r² = 0
Hence the circles are
x² + y² – 2xr₁ – 2yr₁ + r₁2 = 0 ......(1)

x² + y² – 2xr₂ – 2yr₂ + r₂² = 0 .....(2)

As (1) and (2) are orthogonal so
2r₁r₂ + 2r₁r₂ = r₁² + r₂²
4 $\frac{r_{1}}{r_{2}} = {(\frac{r_{1}}{r_{2}})}^{2}$ + 1
$\Rightarrow {(\frac{r_{1}}{r_{2}})}^{2} - 4 {(\frac{r_{1}}{r_{2}})}^{2}$ + 1 = 0
$\Rightarrow \frac{r_{1}}{r_{2}} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{2 \pm 2 \sqrt{3}}{2}$
= 2 + $\sqrt{3}$ or 2 – $\sqrt{3}$ (rejected)

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