Question

If $\frac{3}{5}+\frac{5}{15}+\frac{7}{45}+\frac{9}{135}+\cdots \infty =\frac{a}{b}$ (where $a,b$ are coprime), then the equation of circle with $(a,0)$ and $(0,b)$ as end points of a diameter, is

• A. $x^2+y^2+6x+5y=0$
• B. $x^2+y^2-6x-5y=0$
• C. $x^2+y^2+5x+6y=0$
• D. $x^2+y^2-5x-6y=0$

Answer 1

The correct option is B $x^2+y^2-6x-5y=0$

Let $S=\frac{3}{5}+\frac{5}{15}+\frac{7}{45}+\frac{9}{135}+\cdots \infty$
$\Rightarrow 5S=3\cdot 1+5\cdot \frac{1}{3}+7\cdot \frac{1}{3^2}+9\cdot \frac{1}{3^3}+\cdots \infty -\left(1\right)$
$\Rightarrow \frac{5}{3}S=3\cdot \frac{1}{3}+5\cdot \frac{1}{3^2}+7\cdot \frac{1}{3^3}+\dots \infty -\left(2\right)$

From $\left(1\right)-\left(2\right),$
$\frac{10}{3}S=3+2\cdot \frac{1}{3}+2\cdot \frac{1}{3^2}+2\cdot \frac{1}{3^3}+\cdots \infty$
$\Rightarrow \frac{10}{3}S=3+\frac{2/3}{1-1/3}$
$\Rightarrow S=\frac{6}{5}=\frac{a}{b}$
$\Rightarrow a=6,b=5$
Equation of circle with $(6,0),(0,5)$ as diametric end points is
$(x-6)(x-0)+(y-0)(y-5)=0$
$\Rightarrow x^2+y^2-6x-5y=0$