circle tangent theorem 11-1 if a line is tangent to a circle, then the line is perpendicular to the...
TRANSCRIPT
Tangent
Theorem 11-1
If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangencyAB is tangent to the circle since the segment touches the circle once and intersects with the radius at a 90° angle.
Theorem 11-2:
• If a line in the plane of a circle is perpendicular to a radius at its endpoint on the circle, then the line is tangent to the circle.
Theorem 11-2:
• If a line in the plane of a circle is perpendicular to a radius at its outer endpoint, then the line is tangent to the circle.
• I is perpendicular to radius OB at B
What must be the length of LM for this segment to be tangent line of the circle with center N?
The answer is:
For segment LM to be a tangent it intersect the radius MN at 90°. Therefore triangle LMN would have to be a right triangle and the Pythagorean theorem provides the necessary length for LM to be a tangent.
Theorem 11-3
• The two segments tangent to a circle from a point outside the circle are congruent
AM = BM
CD = CE
Theorem 11-3
• if two tangents are drawn on a circle and they cross, the lengths of the two tangents (from the point where they touch the circle to the point where they cross) will be the same.
Theorem 11-4
Within a circle or in congruent circles
1. Congruent central angles have congruent chords
2. Congruent chords have congruent arcs
3. Congruent arcs have congruent central angles
Theorem 11-5
Within a circle or in congruent circles:
1. Chords equidistant from the center are congruent
2. Congruent chords are equidistant from the center
Theorem 11-5
• When congruent chords are in the same circle, they are equidistant from the center.
Theorem 11-6
• In a circle, a diameter that is perpendicular to a chord bisects the chord and its arcs
Theorem 11-7
• In a circle, a diameter that bisects a chord (that is not a diameter) is perpendicular to the chord
Theorem 11-8
• In a circle, the perpendicular bisector of a chord contains the center of the circle
Theorem 11-9
• The measure of an inscribed angle is half the measure of its intercepted arc.
Inscribed Angle = Intercepted Arc
Corollaries to the Inscribed Angle Theorem
1. Two inscribed angles that intercept the same arc are congruent
2. An angle inscribed in a semicircle is a right angle.
3. The opposite angles of a quadrilateral inscribed in a circle are supplementary
A+C=180ºD+B=180º
140º900
40º
900
A
C
D
B
In any cyclic Quadrilateral opposite corners sum to 1800
Any 4 points on a circle joined to form a quadrilateral
Theorem 11-10
• The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc.
Two secants extend from the same point and intersect the circle as
shown in the diagram below. What is the value of x?
The answer is:
x = ½(140-50) x = ½(90) x = 45°
Find the measure of angle k, where the two secant segments intersect.
The answer is:
K = ½(90-30) K = ½(60) K = 30°
Theorem 11-11
• The measure of an angle formed by two lines that :
1. Intersect inside a circle is half the sum of the measures of the intercepted arcs.
Theorem 11-11
2. Intersect outside a circle is half the difference of the measures of the intercepted arcs.
Theorem 11-11
Theorem 11-11
• Use the theorem for the intersection of a tangent and a secant to find the measure of the angle formed by the intersection of the tangent and the secant.
The answer is:
Two Tangents:
<ABC is formed by two tangents intersecting outside of circle O.
The intercepted arcs are minor arc AC and major arc AC. These two arcs together
comprise the entire circle.
Theorem 11-12
• For a given point circle, the product of the lengths of the two segments from the point to the circle is constant along any line through the point and circle.
(DM+ME)DM = (FN+NE)NE (QK+KR)=SR²
• In the circle below, the chord segments have the following lengths: A= 6, C=3, D=4. Use the theorem for the product of chord segments to find the value of D.
The answer is:
Theorem 11-13
• An equation of a circle with center (h ,k) and radius r is (x-h)² = r² (standard Form)
(x - h)2 + (y - k)2 = r2 (x - 3)2 + (y - 5)2 = (3sqrt(2))r2 (x - 3)2 + (y - 5)2 = 18
(x - h)2 + (y - k)2 = r2
• Angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle. So c is a right angle.
• Definition: A circle is the set of points equidistant from a point C(h,k) called the center. The fixed distance r from the center to any point on the circle is called the radius.The standard equation of a circle with center C(h,k) and radius r is as follows:
• (x - h)2 + (y - k)2 = r2• Example 1: Find the equation of a circle whose center is at (2, - 4)
and radius 5.Solution to Example 1: given (h , k ) = (2 , - 4) and r = 5substitute h, k and r in the standard equation (x - 2)2 + (y - (- 4))2 = 52 (x - 2)2 + (y + 4)2 = 25Go here and set h, k and r parameters into applet and plot the circle. Verify graphically that the equation is that of a circle with the given center and radius.Matched Exercise 1: Find the equation of a circle whose center is at (2 , - 4) and radius 3.Answers.
• Example 2: Find the equation of a circle that has a diameter with the endpoints given by the points A(-1 , 2) and B(3 , 2).Solution to Example 2: The center of the circle is the midpoint of the line segment making the diameter AB. The midpoint formula is used to find the coordinates of the center C of the circle. x coordinate of C = (-1 + 3) /2 = 1 y coordinate of C = (2 + 2) / 2 = 2The radius is half the distance between A and B. r = (1/2) ([3 - (-1)]2 + [2 - 2]2 )1/2 = (1/2)(42 + 02)1/2 = 2The coordinate of C and the radius are used in the standard equation of the circle to obtain the equation: (x - 1)2 + (y - 2)2 = 22 (x - 1)2 + (y - 2)2 = 4 Go here and set h, k and r parameters into applet and plot the circle. Verify graphically that the equation is that of a circle with the diameter as given above.Matched Exercise 2: Find the equation of a circle that has a diameter with the endpoints given by A(0 , -2) and B(0 , 2).Answers.
• Example 3: Find the center and radius of the circle with equation x2 - 4x + y2 - 6y + 9 = 0Solution to Example 3: In order to find the center and the radius of the circle, we first rewrite the given equation into the standard form as given above in the definition. Put all terms with x and x2 together and all terms with y and y2 together using brackets. (x2 - 4x) +( y2 - 6y) + 9 = 0We now complete the square within each bracket.. (x2 - 4x + 4) - 4 + ( y2 - 6y + 9) - 9 + 9 = 0 (x - 2)2 + ( y - 3)2 - 4 - 9 + 9 = 0Simplify and write in standard form (x - 2)2 + ( y - 3)2 = 4 (x - 2)2 + ( y - 3)2 = 22 We now compare this equation and the standard equation to obtain. center at C(h , k) = C(2 , 3) and radius r = 2Matched Exercise 3: Find the center and radius of the circle with equation x2 - 2x + y2 - 8y + 1 = 0Answers. Example 4: Is the point P(3 ,
• 4) inside, outside or on the circle with equation (x + 2)2 + ( y - 3)2 = 9 Solution to Example 4: We first find the distance from the center of the circle to point P. center C at (-2 , 3) radius r = (9)1/2 = 3 distance from C to P = ([3 - (-2)]2 + [4 - 3]2)1/2 = (52 +12)1/2 = (26)1/2Since the distance from C to P is (26)1/2 which approximately equal to 5.1 is greater than the radius r = 3, point P is outside the circle. You can check your answer graphically using this applet.Matched Exercise 4: Is the point P(-1 , -3) inside, outside or on the circle with equation (x - 1)2 + ( y + 3)2 = 4
• Alternate Segment Theorem
