in circle O, perpendicular chords AB and CD intersect at E so that AE=2, eb=12 and CE=4. find the radius of a circle O and the shortest distance from e to the circle
The two chords intersect to form 2 similar triangles, in this case 2 similar right-angled triangles
so we can easily find that ED = 24
by Pythagoras:
AC = √20
and BD = 6√20
Does that give you a start ?
To find the radius of circle O, we can use the relationship between the length of a perpendicular from the center of a circle to a chord and the lengths of the chord segments.
Let's call the center of the circle O, and let r be the radius of the circle. Since AE and CE are perpendicular chords, they intersect at the center of the circle O. The length of AE is given as 2, and EB as 12, so AB = AE + EB = 2 + 12 = 14. Similarly, CD = CE + ED = 4 + ED.
Now, since AB and CD are perpendicular chords that intersect at the center of the circle O, we know that the product of the lengths of the chord segments must be equal. In other words, AE * EB = CE * ED.
Substituting the given values, we have 2 * 12 = 4 * ED. Solving this equation, we find that ED = 6.
Now, since ED is a radius of the circle, we can conclude that r = ED = 6. Therefore, the radius of circle O is 6 units.
To find the shortest distance from E to the circle O, we need to find the length of the perpendicular segment from E to the circle. Let's call this distance x.
By the Pythagorean theorem, we can set up an equation:
x^2 + (r - 2)^2 = r^2
Substituting the value of r=6, we have: x^2 + 4^2 = 6^2
Simplifying this equation, we get: x^2 + 16 = 36
Subtracting 16 from both sides, we have: x^2 = 20
Taking the square root of both sides, we find: x ≈ √20, which simplifies to x ≈ 2√5.
Therefore, the shortest distance from E to the circle O is approximately 2√5 units.