If you have a circle and any random point, you must be curious about where that point is, right? Is the point inside the circle, exactly on the edge of the circle, or completely outside the circle?
This concept is really important because in real life we often need to know the position of an object relative to a circular area. For example, is your house still within the signal range of a tower that's circular in shape, or is an airplane's position still within radar surveillance.
By using circle equations and point coordinates, we can determine the position of that point mathematically and accurately.
To determine the position of a point relative to a circle, we use a concept called power of a point. This is a mathematical way to measure "how far" that point is from the circle.
If we have point A(x0,y0) and a circle with general equation x2+y2+Dx+Ey+F=0, then the power of point A is defined as:
KA=x02+y02+Dx0+Ey0+F
So just substitute the point coordinates into the circle equation, easy right?
The process is quite straightforward. First, we identify the circle equation and coordinates of the point to be checked. Second, substitute the point coordinates into the circle equation to get the power of a point value. Third, look at the sign of the substitution result.
If the result is negative, the point is inside. If zero, the point is on the circle. If positive, the point is outside the circle.
Let's try with a concrete example. Say we have point A(1,−2) and circle x2+y2=25.
Substitute the point coordinates:
KA=12+(−2)2=1+4=5
Since the power of point value KA=5 and for circle x2+y2=25 which has r2=25, then 5<25. So point A(1,−2) is located inside the circle.
For circles in the form x2+y2=r2, we compare the substitution result with r2. If less than r2, point is inside. If equal to r2, point is on the circle. If greater than r2, point is outside.
If the circle is in general form x2+y2+Dx+Ey+F=0, the method is the same. Just substitute the point coordinates into the entire equation and look at the sign of the result.
For example, for point A(1,−2) and circle x2+y2−8x−2y−8=0:
KA=12+(−2)2−8(1)−2(−2)−8
KA=1+4−8+4−8=−7
Since KA=−7<0, then point A(1,−2) is located inside the circle.
