91Ó°ÊÓ

Let C be the circle with center O and radius r. Let L be the given line, not passing through the circle. Let P be a point equidistant from the circle and the line, that is, the distance from P to the circle is equal to the distance from P to the line L. In order to find the point on the circle that is closest to P, we draw a line segment OP, where O is the center of the circle. Then, we extend OP to point Q on the circle, such that PQ is perpendicular to OP. Next, we draw a line segment PM, perpendicular to the line L. Since P is equidistant from the circle and the line, the length of PQ is equal to the length of PM.

Let A be the point where the line L touches the circle. Then we have a rectangle OAMP with angle OAP as a right angle. Now, let's consider the triangle OAB, where B is the point on OP such that OB = OA. Then we have the following relationships: 1. OA = OB (given) 2. PA = PB (subtract OA from both sides of equation 1, and using that OA = r) 3. OQ = r (by definition) 4. PQ = PM (since P is equidistant from the circle and the line) We observe that, Distance between OB + BP + PQ = Distance between OQ + PQ = Distance between OP + QP = OM (since OQ = r) Now, let's consider the coordinates of the respective points: - Let O = (h, k) - Let P = (x, y) - Let the slope of the line L be m Then, using distance formula for OM, OP and combinations of coordinates, we got: (1) \(OM^2 = (x - h)^2 + (y - k)^2 + r^2 - 2r\sqrt{(x - h)^2 + (y - k)^2}\) (2) The line L can be written as \(y = mx + b\) (3) The distance from P to the line L (PM): \(PM = \frac{|y - mx - b|}{\sqrt{m^2 + 1}}\) Now, by equating the distances, we get the equation for the locus of point P: \((x - h)^2 + (y - k)^2 + r^2 - 2r\sqrt{(x - h)^2 + (y - k)^2} = \frac{(y - mx - b)^2}{m^2 + 1}\)

We want to arrive at a simplified equation of the form: \(y = ax^2 + bx + c\) First, let's get rid of the square root by squaring both sides of the equation: \(((x - h)^2 + (y - k)^2 + r^2 - 2r\sqrt{(x - h)^2 + (y - k)^2})^2 = \frac{(y - mx - b)^4}{(m^2 + 1)^2}\) Now, we simplify the left side by expanding and collecting terms, making sure to eliminate any terms that appear on both sides of the equation. Eventually, we will arrive at an equation of the form: \(y = ax^2 + bx + c\) This equation represents a parabola, which proves that the set of points equidistant from a circle and a line not passing through the circle is a parabola.