In circle O, perpendicular chords AB and CD intersect at E so that AE= 2, EB= 12 and CE= 4. Find the radius of circle O and the shortest distance from E to the circle.
If wet up coordinates so that O directly below the center of AB.
and A is at x = -7 and B is at x=7
then C is at (-5,4)
segment AC has slope 2
The radius R is the perpendicular bisector of AC, so it has slope -1/2
If F is the midpoint of AC then F is at (-6,3), making O = (0,-1)
so R = 3√5
If G is the closest point on the circle to E, then G is where the circle intersects OE extended. That is, where
x^2 + (y-1)^2 = 45
y-1 = -1/5 (x+5)
Now just find the distance EG where E = (-5,1)
To solve this problem, we can use the property that the perpendicular bisectors of the chords in a circle intersect at the center of the circle.
Let's start by finding the center of circle O. Draw the perpendicular bisectors of chords AB and CD. Label the points where the perpendicular bisectors intersect as F.
Since AE = 2 and EB = 12, the length of AB is AE + EB = 2 + 12 = 14. Similarly, since CE = 4, the length of CD is CE + EB = 4 + 12 = 16. Therefore, AB and CD are chords of lengths 14 and 16, respectively.
Since the perpendicular bisectors of AB and CD intersect at point E, this means that point E is the center of circle O.
Next, let's find the radius of the circle. From the center E, draw a radius to either point A or B. Since AE is given as 2, the radius of circle O is also 2 units.
Now, let's find the shortest distance from point E to the circle. Recall that a radius of a circle is always perpendicular to the circle at the point of intersection. Therefore, the shortest distance from point E to the circle is the length of the radius, which is 2 units.
In summary, the radius of circle O is 2 units, and the shortest distance from point E to the circle is also 2 units.