Then we know that OC bisects AB, and BC is 3 units long. Then we can use the HL Congruency to show that the triangles are congruent and then using CPCTC that the AC and BC are congruent.
We need to use the fact that the radius is the hypotenuse of both right triangles. In geometry, we cannot use the fact that it seems like about half of 6. HOWEVER, the point with this question is that we don't know without using some other geometry that the small leg is actually 3. Then we learn that 3, 4, 5 is a Pythagorean triplet like 12, 13, 5 and 24, 7, 25 and 6, 8, 10 The reason we can use the 3, 4, 5-triangle AFTER we know for sure that BC = 3 is that we know from using the Pythagorean Theorem (once or dozens of times) that our result will be 4 IF the hypotenuse is 5 and the other leg is 3. Here we DON'T know that the small leg is 3 at first. Xx 1 + yy 1 – 3(x + x 1) –2(y + y 1) + 3 = 0.Yes, you can always do that if you encounter a right triangle with a hypotenuse of 5 and one leg measuring 3. The equation of the chord of contact of P with respect to the given circle is Hence the equation of the required circle is (x – 3) 2 + (y – 4) 2 = (62/13) 2.įind the co-ordinates of the point from which tangents are drawn to the circle x 2 + y 2– 6x – 4y + 3 = 0 such that the mid-point of its chord of contact is (1, 1). So, OB is a perpendicular bisector of PQ.įind the equation of the circle whose center is (3, 4) and which touches the line 5x + 12y = 1. If in the given circle with center O, the length of PQ is 10cm, then determine PA.Īs it is visible in the figure, OB is perpendicular to PQ. Result: The arcs intercepted by two congruent chords are congruent.Ĭonverse: If two arcs are congruent then their corresponding chords are congruent. Chords equidistant from the center of a circle are congruent. Result: If two chords are congruent then they are equidistant from the center. OB is the perpendicular bisector of the chord RS and it passes through the center of the circle. In the above circle, OA is the perpendicular bisector of the chord PQ and it passes through the center of the circle. In the above circle, if the radius OB is perpendicular to the chord PQ then PA = AQ.Ĭonverse: The perpendicular bisector of a chord passes through the center of a circle.
Result: A radius or diameter that is perpendicular to a chord divides the chord into two equal parts and vice versa. Then the length of the tangent segment squared is equal to the product of the secant segment and its external segment. Result: Consider a circle with a tangent segment and a secant segment. The product of the segments of one chord is equal to the product of the segments of the other chord. Result: In case of intersection of two chords in the same circle, each chord gets divided into two segments by the other chord. We discuss a few of them here as they often prove helpful in solving various questions. There are various important results based on the chord of a circle. Where r is the radius of the circle and d is the perpendicular distance of the center of the circle to the chord.
In case, you are given the radius and the distance of the center of circle to the chord, you can apply this formula: Where r is the radius of the circle and c is the angle subtended at the center When the radius and a central angle are given, the length of the chord can be computed using the given formula: There are two ways to find the length of the chord depending on what information is provided in the equation: Methods of finding the length of the chord The equation of the chord of the circle x 2 + y 2 + 2gx + 2fy +c=0 with M(x 1, y 1) as the midpoint of the chord is given by: Note that the end points of such a line segment lie on the circle. Concepts of Physics by HC Verma for JEEĪ line that links two points on a circle is called a chord.IIT JEE Coaching For Foundation Classes.
0 kommentar(er)
