Question

Prove that circles x²+y²=ax and x²+y²=by are orthogonal.

Answer 1

Answer:

The given curves are orthogonal

Step-by-step explanation:

Prove that circles x²+y²=ax and x²+y²=by are orthogonal.

$\text{Let }(x_1,y_1)\text{ be the point of intersection of given two curves}$

Then,

$x_1^2+y_1^2=ax_1$.......(1)

$x_1^2+y_1^2=ay_1$.......(2)

$x^2+y^2=ax$

Differentiate with respect to x

$2x+2y\:\frac{dy}{dx}=a$

$\mplies\:\frac{dy}{dx}=\frac{a-2x}{2y}$

$\text{slope of tangent }m_1=(\frac{dy}{dx})(x_1,y_1)$

$\implies\:m_1=\frac{a-2x_1}{2y_1}$

$x^2+y^2=ay$

Differentiate with respect to x

$2x+2y\:\frac{dy}{dx}=a\frac{dy}{dx}$

$2y\:\frac{dy}{dx}-a\frac{dy}{dx}=-2x$

$\frac{dy}{dx}(2y-a)=-2x$

$\frac{dy}{dx}=\frac{-2x}{2y-a}$

$\text{slope of tangent }m_2=(\frac{dy}{dx})(x_1,y_1)$

$\implies\:m_2=\frac{-2x_1}{2y_1-a}$

Now,

$m_1\times\:m_2$

$=\frac{a-2x_1}{2y_1}\times\:\frac{-2x_1}{2y_1-a}$

$=\frac{a-2x_1}{y_1}\times\:\frac{-x_1}{2y_1-a}$

$=\frac{-ax_1+2x_1^2}{2y_1^2-ay_1}$

$=\frac{x_1^2+(x_1^2-ax_1)}{y_1^2+(y_1^2-ay_1)}\:\text{ using(1) and (2)}$

$=\frac{x_1^2-y_1^2}{y_1^2-x_1^2}$

$=\frac{x_1^2-y_1^2}{-(x_1^2-y_1^2)}$

$=-1$

$\implies\:\boxed{m_1\times\:m_2=-1}$

$\text{Hence, the given curves are orthogonal}$