Question

The equation of the tangent at the point $(0,0)$ to the circle where circle makes intercepts of length $2a$ and $2b$ units on the coordinates axes, is (are)-

• A. $ax+by=0$
• B. $ax-by=0$
• C. $x=y$
• D. $bx+ay=ab$

Answer 1

Answer: (A)

$ax+by=0$

Let the equation of the circle be,

${(x-a)}^2{+(y-b)}^2=r^2⟶\left(1\right)$

It makes an intercept of length $2a$ and $2b$ units on the coordinates axes, i.e. it passes through $(2a,0)$ and $(0,2b)$.

$\therefore$ equation of the circle becomes,

$a^2+b^2=r^2$, passing through $(2a,0)$ and $(0,2b)$

and equation $\left(1\right)$ becomes,

$x^2-2ax+a^2+y^2-2by+b^2=r^2\Rightarrow x^2-2ax+y^2-2by+a^2+b^2=r^2\Rightarrow x^2-2ax+y^2-2by+r^2=r^2\Rightarrow x^2+y^2-2ax-2by=0$

Since, $(0,0)$ passes through above the circle,

Thus, centre of the circle=$(a,b)$

$\therefore$ Equation of line passing through $(0,0)$ and $(a,b)$ is

$ay=bx\Rightarrow bx-ay=0⟶\left(2\right)$

Equation of the line perpendicular to the equation $\left(2\right)$ is,

$ax+by+k=0⟶\left(3\right)$

Since, it passes through $(0,0)$,

So, $a\times 0+b\times 0+k=0\Rightarrow k=0$

$\therefore$ equation $\left(3\right)$ becomes,

$ax+by=0$

Hence, this is the required equation of the tangent at the point $(0,0)$ to the given circle.