# If the lines represented by the equation x^2+y^2=c^2((bx+ay)/(ab))^2 form a right angle then prove that:1/a^2+1/b^2+1/c^2=3/c^2?

## $\frac{1}{a} ^ 2 + \frac{1}{b} ^ 2 + \frac{1}{c} ^ 2 = \frac{3}{c} ^ 2$

Dec 10, 2017

See below.

#### Explanation:

The product of lines is homogeneous in $x , y$ so the lines should be of the form

${L}_{1} \to y - m x = 0$
${L}_{2} \to x + m y = 0$

then

$\left({a}^{2} {b}^{2} - {b}^{2} {c}^{2}\right) {x}^{2} - 2 a b {c}^{2} x y + \left({a}^{2} {b}^{2} - {a}^{2} {c}^{2}\right) {y}^{2} = \left(y - m x\right) \left(x + m y\right)$

grouping variables gives us

{(a^2 b^2 - a^2 c^2 - m = 0), ( m^2- 2 a b c^2 =1), (a^2 b^2 - b^2 c^2 + m=0):}

From the first we have

$\frac{1}{c} ^ 2 - \frac{1}{b} ^ 2 = \frac{m}{{a}^{2} {b}^{2} {c}^{2}}$

from the third we have

$\frac{1}{c} ^ 2 - \frac{1}{a} ^ 2 = - \frac{m}{{a}^{2} {b}^{2} {c}^{2}}$

adding those last equations we have

$\frac{2}{c} ^ 2 - \frac{1}{a} ^ 2 - \frac{1}{b} ^ 2 = 0$ now adding $\frac{1}{c} ^ 2$ to both sides

$\frac{3}{c} ^ 2 - \frac{1}{a} ^ 2 - \frac{1}{b} ^ 2 = \frac{1}{c} ^ 2$ or

$\frac{1}{a} ^ 2 + \frac{1}{b} ^ 2 + \frac{1}{c} ^ 2 = \frac{3}{c} ^ 2$