Given of course that both come in a TO-220 case. I wouldn't think so, right? I try to make it out from the K/W ratings but they seem to vary wildly from manufacturer to manufacturer. Anybody knows the answer?


Question: Intersection points of two circles.
Answer: The two given equations, only the first of which is a circle, intersect at infinitely many points, not at just points $A, B$, and those points of intersection lie along a common line. It is never possible that the points of intersection of two circles are all the points on any one line. 

We can see this by solving for the points of intersection of the two equations: Since the left-hand sides of each equation is equal to zero, the left-hand sides are equal to one another. This gives us 

$$\begin{align} x^2+y^2+4x-10y+20 &amp; = x^2+y^2-4x-2y+12\\ \\  \iff 8x-8y+8&amp; =0 \\ \\ \iff y&amp; =x+1.\end{align}$$

So if there were a valid intersection, the "intersection" would consist of in this line. Can you see why this can not happen, given the first equation is the equation of a circle? If we had the equations of two circles, they would intersect at nowhere, or at one point (if they were tangent), or at 2 points (overlapping but not concurrent), or else at infinitely many points (if they represented the same circle.) None of those possibilities is met by these two equations. So they cannot define circles. The first does in fact define a circle (graph it to see). The second doesn't (no object has a radius of $\sqrt{-7}$, which occurs with the second equation if you complete the square to obtain the center and radius a circle would have, if it in were a circle.

So I suspect at least the second circle was either mistyped, or is a misprint in the exercises you are working on.