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Show tnal tne point where OP intersects @ is the unique closest point t0 P on €. 14G) Suppose & = (O.r) is a circle and P is & point in the exterior o...

Question

Show tnal tne point where OP intersects @ is the unique closest point t0 P on €. 14G) Suppose & = (O.r) is a circle and P is & point in the exterior of € , Show that the point- where OP intersects € is the unique closest point to P on €.

Show tnal tne point where OP intersects @ is the unique closest point t0 P on €. 14G) Suppose & = (O.r) is a circle and P is & point in the exterior of € , Show that the point- where OP intersects € is the unique closest point to P on €.



Answers

Consider a circle $C$ and a point $P$ exterior to the circle. Let line segment $P T$ be tangent to $C$ at $T,$ and let the line through $P$ and the center of $C$ intersect $C$ at $M$ and $N .$ Show that $(P M)(P N)=(P T)^{2}$.

The distance between the two points P X one Y one and r X two Y two is the distance P. R. Equals the square root of x one minus x. Two squared plus Y one minus Y to square. This is the distance formula we're going to find the closest point on the line. Y equals two X. Which is graft right here. That's that line there. And the .17 which we've plotted right there. It's important to remember the shortest distance between two points or in this case the point and the line is going to be a straight line. We're just gonna go right here. If I go any other direction my line is gonna be longer. And when I look at this distance right here, what we're actually looking for is what is this point on the line? And looking at that, if we follow it straight down, We're going to see it's still at seven. So are y. value is still seven And it's why equals two times x. And we just said seven equals 2. X Divide both sides by two and we get 3.5 is X. So the point and the line, The closest point on that line is at 3.5:07

Let it be if B and zero and M x and y on direct Lex de equal be no, if I am, will be equal to root Ex Finance P Square plus voice square so if M will be equal to G plus x so root x minus. P square plus y square equal B plus x so x minus B all square plus y squared equal p plus x all the square. So after submitting cation, we have X square minus two p x plus p square plus voice square equals P square. Last two p x lost exasperation. So we have or square equal four p x

Mhm. Mhm. So here what we have is that um there's a set of points equidistant from a circle and a line outside the circle is a problem. Yeah. So if we have the set of points from a circle so we'll take the circle X squared. Yeah, that's why I spread equals um four and then we know that it's going to be set up point equidistant from a circle and a line outside of the circle. Um So if we have a line outside of the circle such as X equals three, X equals y equals three. Yeah, we see that a problem. Ah What perfectly balance itself around this, where it be located here and then as we move this way it would kind of rotate around the circle but also stay within a distance from the line as it would right here. So we can see this visually if we picture a problems stretching out and maintaining equal distance from the the circle and a line that doesn't cross the circle.


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