Question

How do you find the equation of tangent of circle ${x^2} + {y^2} = 25$ at the point $( - 3,4)$?

Answer 1

Hint: Here the circle is not a function; we will split it into two semi-circles and solve for the semi-circle which is in the quadrant of the given point and find its slope by taking the derivative of the equation, and then use the point-slope formula to get the final answer.

Formula used: $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
$\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}$

Complete step-by-step solution:
We have the given equation of the circle as:
$ \Rightarrow {x^2} + {y^2} = 25$
On rearranging the terms, we get:
$ \Rightarrow {y^2} = 25 - {x^2}$
On taking the square root on both the sides, we get:
$ \Rightarrow y = \pm \sqrt {25 - {x^2}} $
Now since the point $( - 3,4)$ is a point in the second quadrant, the $y$ intercept will be positive, therefore the function we need is:
$ \Rightarrow y = + \sqrt {25 - {x^2}} $
Now to get the slope, we will find the derivative of this equation, on differentiating, we get:
$ \Rightarrow y' = \dfrac{d}{{dx}}\sqrt {25 - {x^2}} $
Now the above term has $2$functions, therefore we will use the chain rule which is:
$F'(x) = f'(g(x))g'(x)$
Here we use the formula $\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}$ therefore:
$ \Rightarrow y' = \dfrac{1}{{2\sqrt {25 - {x^2}} }}\dfrac{d}{{dx}}(25 - {x^2})$
Now we know that $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}k = 0$ therefore:
$ \Rightarrow y' = \dfrac{{ - 2x}}{{2\sqrt {25 - {x^2}} }}$
On simplifying the equation, we get:
$ \Rightarrow y' = \dfrac{{ - x}}{{\sqrt {25 - {x^2}} }}$
Now the slope at the point $( - 3,4)$ is:
$ \Rightarrow y'( - 3) = \dfrac{{ - 3}}{{\sqrt {25 - {3^2}} }}$
On simplifying we get:
$ \Rightarrow y'( - 3) = \dfrac{{ - ( - 3)}}{{\sqrt {25 - 9} }}$
On simplifying and taking the square root, we get:
$ \Rightarrow y'( - 3) = \dfrac{3}{4}$
Therefore, the slope $m = \dfrac{3}{4}$.
Now using the point slope formula $y - {y_1} = m(x - {x_1})$, we get:
$ \Rightarrow y - 4 = \dfrac{3}{4}(x - ( - 3))$

On simplifying and rearranging the terms, we get:
$ \Rightarrow y = \dfrac{3}{4}x + \dfrac{{25}}{4}$, which is the required tangent line to the point $( - 3,4)$.

Note: It is to be remembered that taking the derivative for the equation of a line or a curve, the slope of the line is found.
A tangent is a line which touches a curved line at a specific point; the curve can be a circle, ellipse etc.