Question

If the centre of a circle is $(-6,8)$ and it passes through the origin, then equation to its tangent at the origin , is

• A. $2y=x$
• B. $4y=3x$
• C. $3y=4x$
• D. $3x+4y=0$

Answer 1

The correct option is A $2y=x$

Center of a circle is given as $C(-6,8)$

The circle is passing through origin $O(0,0)$

Thus, as shown in figure, OC will be radius of the circle.

By distance formula, $d\left(OC\right)=\sqrt{{(-6-0)}^2-{(8-0)}^2}$

$\therefore d\left(OC\right)=\sqrt{36+64}=\sqrt{100}$

$\therefore d\left(OC\right)=r=10$

Now, draw a tangent to a circle through origin as shown in figure.

By property of tangent-radius, radius OC will be perpendicular to tangent through O.

Let, $m_1$ = Slope of radius OC

$m_2$ = Slope of tangent

$\therefore m_1\times m_2=-1$ Equation (1)

Now, equation of slope of radius OC is given as,

$m_1=\frac{y_O-y_C}{x_O-x_C}$

$\therefore m_1=\frac{0-8}{0-(-6)}=\frac{-8}{6}$

$\therefore m_1=-\frac{4}{3}$

Thus, from equation (1),

$-\frac{4}{3}\times m_2=-1$

$\therefore m_2=\frac{3}{4}$

Now, Equation of tangent is given by two point form as,

$y-y_1=m_2(x-x_1)$

Here, $x_1$ and $y_1$ are coordinates of origin as tangent is passing through origin.

$\therefore y-0=\frac{3}{4}(x-0)$

$\therefore y=\frac{3}{4}x$

$\therefore 4y=3x$

Thus, answer is option (B)