Before addressing my issues, below is the question from a past examination paper along with a diagram I dre in order to facilitate readers.
3(a) A circle has center $C(5, 8)$ and radius $\sqrt{29}$
(ii) The circle cuts the y-axis at $R$ and $Q$, where $Q$ is above $R$. Find the coordinates of $Q$ and $R$, and find the equation of the tangent at $Q$.
Diagram
First thing I did (well, it was requested) was find the equation of this circle, which turned out to be $x^2-10x+y^2-16y+60=0$.
Now, not only from the graph shown above but also the one I drew on paper I am able to find out that the points for $Q$ and $R$ are $(0,10)$ and $(0,6)$ respectively. However, that is not what the examiner would like the student to do (granted I got the answer anyway and still think I shouldn't be penalized for it).
Given that I have no information apart form the fact that triangle $CQR$ is an isosceles triangle with two equal sides having the length $\sqrt29$.
With this information alone I'm not quite capable (yet) to proceed with the question. How would one proceed?
Step 1.
Equation of circle: $x^2−10x+y^2−16y+60=0$
Step 2.
For points Q and R substitute values of $x=0$, and $y=y$ into the circle equation.
That will give you $y^2-16y+60=0$
Step 3.
Factorize this to get $(y-10)(y-6)=0$.
Therefore $y=6$ and $y=10$.
Step 4.
So, we have the two point $(0,10)$ and $(0,6)$.
Step 5.
Using the point $(5,8)$ and $(0,10)$ ,find the slope of the line. Using the equation $\dfrac{y1-y2}{x1-x2}$.
Therefore slope is $\frac{-2} {5}$.
Step 6.
Since the tangent is perpendicular to the line $CQ$. The slope of the tangent is $\frac{5}{2}$.
$\text{Since slope 1*slope 2= -1}$.
Step 7.
Therefore we can form the equation using the slope and the point $(0,10)$.
Step 8.
The equation is $(y-10)=\frac{5}{2}(x-0)$. Which is $2y-20=5x$.