# The equation of the tangents to the circle x^2 + y^2 = 25 with slope 2 is: A y = 2x ± √5 B y = 2x ± 2√3 C y = 2x ± 3√5 D y = 2x ± 5√5

## Question

The equation of the tangents to the circle $x^{2}+y^{2}=25$ with slope 2 is

A $y=2 x \pm \sqrt{5}$

B $y=2 x \pm 2 \sqrt{3}$

C $y=2 x \pm 3 \sqrt{5}$

D $y=2 x \pm 5 \sqrt{5}$

✨ Free AI Tools for You

Powered by Chatterbot AI

## Answer

To solve the problem of finding the equation of tangents to the circle $x^2 + y^2 = 25$ with a given slope of 2, we can use the general equation of the tangent to a circle, which incorporates the slope of the tangent.

Given:

- Equation of the circle: $x^2 + y^2 = 25$
- Slope of the tangent, $m = 2$

First, identify the radius of the circle:

- The circle's equation can be written as $x^2 + y^2 = a^2$, where $a = 5$.

The equation of the tangent to the circle with slope (m) is: $$ y = mx + c $$

We need to determine the value of $c$. The condition for this line to be a tangent to the circle is that the perpendicular distance from the center of the circle (0, 0) to the line $y = mx + c$ equals the radius of the circle. Mathematically, this condition can be expressed as: $$ \frac{|c|}{\sqrt{1 + m^2}} = a $$

Given:

- $m = 2$
- $a = 5$

Substitute these values into the equation: $$ \frac{|c|}{\sqrt{1 + 2^2}} = 5 $$ $$ \frac{|c|}{\sqrt{1 + 4}} = 5 $$ $$ \frac{|c|}{\sqrt{5}} = 5 $$ $$ |c| = 5 \sqrt{5} $$

Thus: $$ c = \pm 5 \sqrt{5} $$

The equations of the tangents to the circle with slope 2 are: $$ y = 2x + 5\sqrt{5} \quad \text{and} \quad y = 2x - 5\sqrt{5} $$

Therefore, the correct answer is:
**Option D**: $y = 2x \pm 5\sqrt{5}$

India's 1^{st} AI Doubt Solver for CBSE, JEE, and NEET

and then it's just ₹212 a month