Question

The equation of the tangent to the circle $x^2+y^2=25$ which is inclined at ${60}^{\circ }$ angle with x-axis, will be

• A. $y=\sqrt{3}x\pm 10$
• B. $y=\sqrt{3}x\pm 2$
• C. $\sqrt{3}y=x\pm 10$
• D. None of these

Answer 1

Answer: (A)

$y=\sqrt{3}x\pm 10$

Let the equation of the tangent be $y=mx+c$

Since inclination =60 degrees

$\Rightarrow m=\tan 60=\sqrt{3}$

So, the equ. of the tangent will be $y=\sqrt{3}x+c$

Now putting $y=\sqrt{3}x+c$ in given equation of circle, we get

$\Rightarrow x^2+(\sqrt{3}x+c{)}^2=25$

$\Rightarrow 4x^2+2\sqrt{3}cx+c^2-25=0$

Now since we need to find value of $c$ for equ. to become tangent

$\Rightarrow$The above quadratic equ. must have real and equal roots

$\Rightarrow D=0$

$\Rightarrow (2\sqrt{3}c{)}^2-4(4\left)\right(c^2-25)=0$

$\Rightarrow 4c^2=400$

$\Rightarrow c=\pm 10$

So, the equation of the tangent is $y=\sqrt{3}x\pm 10$

Hence, the answer is $A$