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In circle O, the chord Mn / / EF, P is a point on circle O. the radius of circle O is known to be 10cm, Mn = 12cm, EP = 16cm. Find the distance between chord Mn and ef
Take the midpoint P and Q of Mn and ef
∴OP⊥MN,OQ⊥EF,MP=MN/2=6, EQ=EF/2=8
∵MN‖EF
The three points of OPQ are collinear, and the distance between Mn and EF is PQ
∵OP=√(OM^2-MP^2)=8, OQ=√(OE^2-EQ^2)=6
ν PQ = OP + OQ = 14 (Mn, EF on the opposite side)
Or PQ = op-oq = 2 (Mn, EF on the same side)
The distance between Mn and EF is 14 or 2
AB is the diameter of circle O, and the chord CD is perpendicular to P. if AP: Pb = 1:4, CD = 8, find the length of diameter ab The picture is drawn according to the title!
If AP: Pb = 1:4, let AP = m, Pb = 4m, so OC = OA = ob = (OA + OB) / 2 = (AB) / 2 = 5m / 2PO = oa-ap = 5m / 2-m = 3m / 2 because the chord CD is perpendicular to P. according to the Pythagorean theorem, CP = √ (OC ^ 2-op ^ 2) = 2mcd = 2cp = 4m
AB is the diameter of circle O, CD is the chord of circle O, and CD is perpendicular to ab. if AP ratio Pb = 1:4 and CD = 8, what is ab?
Draw a picture first
Then let Pb = x, then AP = 4x; ab = 5x, O is the midpoint, so OP = 1.5x, C0 is the radius, so C0 = 2.5x, CP = 4 (because AB bisects CD ah) and then formulate the equation
（1.5X）^2+4*4=(2.5X)^2
X = 2 then AB = 5x = 10
If the radius of center O is OA = 5cm, chord AB = 8cm, and C is the midpoint of AB, then what is the length of OC?
∵ C is the midpoint of ab
∴AC=AB=4
According to the vertical diameter theorem
OC is perpendicular to ab
In RT △ OAC
OA=5
AC=4
So OC = 3
The length of OC is 3cm
As shown in the figure, if the radius of ⊙ o is OA = 6, chord AB = 8, and P is the moving point on AB, then the shortest distance from point P to center O is______ .
When OD ⊥ AB passes through point O to point D, the distance from point P to center O is the shortest when point P is perpendicular to point D,
∵OD⊥AB，AB=8，
∴AD=1
2AB=1
2×8=4，
In RT △ AOD,
∵OA=6，AD=4，
∴OD=
OA2−AD2=
62−42=2
5，
The shortest distance from point P to center O is 2
5．
So the answer is: 2
5．
As shown in the figure, the radius OA of circle O is 13, and point P is the moving point on chord ab. the shortest distance from point P to center O is 5, then chord AB is equal to () cm To process
OP is the shortest when it is perpendicular to ab,
Because OA = ob = 13
So PA = Pb = radical (13 ^ 2-5 ^ 2) = 12
AB=PA+PB=24
Given that the radius of circle O is 5, the length of chord AB is 8, and the chord BC ∥ OA, find the length of AC
Connect OC, because BC ‖ OA, so ∠ OAB = ∠ ABC, ∠ AOC = ∠ OCB
Because ∠ OAB = ∠ oba, so ∠ OBC = 2 ∠ oba
cos∠OBC=cos∠OBA^2-sin∠OBA^2=(4/5)^2-(3/5)^2=7/25
In isosceles triangle OCB, ∠ OCB = ∠ OBC
So cos ∠ AOC = 7 / 25
According to the cosine theorem
AC^2=OA^2+OC^2-2×OA×OC×cos∠AOC
AC^2=25+25-2×5×5×7/25
AC=6
As shown in the figure, if the radius of circle O is OA = 5, point P is the moving point on chord AB, and the shortest distance between point P and center O is 3, then chord AB is equal to () cm
The radius of circle O is OA = 5, point P is the moving point on chord AB, and the shortest distance between point P and center O is 3,
Half of string AB = (5 ^ 2-3 ^ 2) ^ 0.5 = 4
String AB = 8
It is known that the diameter ab of the circle O intersects with e at the chord CD. We know that AE = 2, be = 6, ∠ DEB = 60 ° and find the length of CD
AE = 2, be = 6, so od = 1 / 2 * (AE + be) = 4, OE = oa-ae = 4-2 = 2
In △ OED, from the sine theorem
OE/sinD=OD/sin60°
2/sinD=4/sin60°
Sind = √ 3 / 4
cosD=√(1-3/16)=√13/4=1/2*CD/OD=CD/8
So CD = 2
Given that the diameter ab of circle O intersects the chord CD at point E, AE = 1cm, be = 5cm, angle DEB = 60 degrees, then CD =?
Because AE = 1cm, be = 5cm
So OE = 2cm
Oh through O is perpendicular to CD and H
So ch = DH
Because ∠ DEB = 60 °, EHO = 90 °
So eh = 1, oh = radical 3
Linking Co
Because CO is the radius of circle o
So co = 3cm
Because Oh = root 3cm
So ch = radical 6cm (Pythagorean theorem)
So CD = 2CD = 2 root sign 6
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