Answer
$(y + 3) = \frac{3}{4} (x - 4)$
Work Step by Step
The question asks you to write the equation for a line tangent to the circle defined by the equation $\left(x-1\right)^2+\left(y-1\right)^2=25$ at the point $(4, -3)$.
To write the equation for a tangent line, we will use point-slope form, which is written as $(y-y_{1})=m(x-x_{1})$, where $y_{1}$ the y-value of a point on our line, $x_{1}$ is the corresponding x-value of this same point on our line, and $m$ is the slope of the line.
Fortunately, the question has already supplied us with a point, $(4,-3)$, so we already know that $y_{1}=-3$, and $x_{1}=4$. This means that we only have to find $m$, the slope of the line. To find $m$, we have to differentiate the original function. Because the supplied equation (which is a circle) is defined in both the $x$ and $y$ variables, we have to use implicit differentiation. We'll differentiate the equation with respect to $x$, which means that the derivative of $x=1$, and the derivative of $y=y'$. The process will also require us to use the power rule and the chain rule. It goes like this:
$2(x-1)(1)+2(y-1)(y')=0$
$2(y-1)(y')=-2(x-1)$
$y'=\frac{-2(x-1)}{2(y-1)}$
$y'=\frac{1-x}{y-1}$
Now, to get $m$, we have to plug in the $x$ and $y$ values that the question gave us, which are $4$ and $-3$ respectively. That goes like this:
$\frac{1-(4)}{(-3)-1}=\frac{-3}{-4}=\frac{3}{4}$
Now that we know $y_{1}=-3$, $x_{1}=4$, and $m=\frac{3}{4}$, we can plug these values into the point-slope equation that we found earlier. That will give us our final answer, which is:
$(y + 3) = \frac{3}{4} (x - 4)$
We can check our answer by graphing both the original equation and the equation for our tangent line. The attached image demonstrates this.
