The equation of the circle, which touches both the axes and the line 4x 3y 6 in the first quadrant and lies below it, is
Answers
Let’s get some definitions straight first:
Because the circle is touching both axes from inside the third quadrant, and since a circle is a set of points equidistant from one central point, we know that the origin of the circle in question lies on the line y=x. This means that each coordinate of the origin point will be equivalent, and also that the circle’s radius (and thus distance to the axes) will be equivalent to the absolute value of any of the coordinates.
We know that a line tangent to a circle will always be perpendicular to a line that passes through both the origin point and the tangential point.
Thus, we need to find a value, let’s say r, that is equivalent to both the distance from the line to the origin point and from the axes to the origin point. Trouble is, we don’t know where exactly that origin point is, so we have a lot of unknowns. Therefore, since I don’t start calculus until school starts (if that would even help), we must clear up these unknowns.
Before anything else, we can rewrite the equation of the line in slope-intercept form so we can work with it easily. This gives us y=34x+2. Since the line from a tangential point and a circle’s origin point is perpendicular to the tangent, we should know that the line y=−43x+b is this line’s perpendicular counterpart, b being the currently unknown y-intercept.
Our final equation, as of now, remains d=r, d being the distance from the tangential point to the origin of the circle and r being the radius of the circle. We have nothing to help us find rthus far, so our best bet is to findd.
Consider the formula for the distance between two points, . The tangential point will be where the given line and the perpendicular line intersect, found by setting the two equations equal to each other and digging it out, resulting in . Note that is still unknown. This will do for .
We can find the same way, except this time finding the intersection between the origin point at the perpendicular line. The origin point lies on , so we can use this to find that . This is also , conveniently.
For we can just plug into the equation for the line you gave in the question, simplifying out to .
can be , since we know that the coordinates of the origin point will be negative and that the radius will always be positive.
So our equation comes out to be:
EDIT: The right side would actually be , oops.
Our only unknown left is , the y-intercept of the line perpendicular to the one you gave us. However, when plugged in to either of the expressions above, we also find the radius of the circle. We could find it analytically, but I’d rather just make up a graph and see where they intersect:
As you can see, , and so .
That’s all we need to make our circle, as we can simply position the circle so that it is tangential to the axes. Our equation comes out to .
Of course, it’s probably just easier to use the formulas, but someone wise once said that you should only rely on formulas that you understand how to derive.