algebra-precalculuscircles
For the circle $x^2+y^2+6x-4y+3=0$ find
a) The center and radius
b) The equation of the tangent line at the point $(-2,5)$
Now, I solved a) and got the equation
$$(x+3)^2+(y-2)^2=10$$ with center $=(-3,2)$ and radius $=\sqrt{10}$
Now, I've never learned about the tangent of a circle, but I think that it's a line that touches the outer end of a circle. But I'm not 100% on that. So if anyone can help me out with this, that would be very beneficial. And please do not solve this for me.
Here's how I was taught to get tangents to conic sections in high school, half a century ago. If you want to get the tangent at the point $(x_0,y_0)$, you take the equation of the conic, and replace "half" of the $x$'s by $x_0$ and "half" of the $y$'s by $y_0$.
In this case, the equation of the curve is $(x-h)^2+(y-k)^2=r^2$, so the tangent at $(x_0,y_0)$ will be $(x_0-h)(x-h) + (y_0-k)(y-k)=r^2$.
A little algebra, using the fact that $x_0^2 + y_0^2 = r^2$, shows that this can also be written in the form $(x_0 -h)(x-x_0) + (y_0 -k)(y-y_0) = 0$.
This works with any conic section curve, and it gives you tangent planes of quadric surfaces, too.
Of course, you can apply the same process even if the point $(x_0,y_0)$ doesn't lie on the curve. In this case, the line you get is called the polar line of the point. It just happens that, for any point lying on the curve, the polar line is the tangent at the point.
I'm curious -- do they still teach this stuff?
We have $(R-r)^2 = (a-r)^2 +b^2$ and hence $$r = \frac{R^2-a^2-b^2}{2(R-a)}$$
Best Answer
The line tangent to the circle at $(-2,5)$ is the straight line through $(-2,5)$ that is perpendicular to the radius of the circle that runs from the centre of the circle at $(-3,2)$ to the point $(-2,5)$. Find the slope of that radius, use that to get the slope of the tangent line, and you’re on your way.
Edit: I just corrected the $x$-coordinate of the centre. Yours has the wrong sign: $x+3=x-(-3)$, so the correct $x$-coordinate is $-3$.