Circle Theorems: Understanding Tangents and Solving Problems
Explore the concepts of tangents from a point, Pythagoras' theorem in circle problem-solving, and the parts of a circle. Discover the properties of tangents from an external point and how to find their lengths in relation to the circle's radius.
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19 April 2025 Circle theorems: Tangents from a point LO: To use Pythagoras theorem to solve problems from circles. www.mathssupport.org
Naming the parts of a circle A circle is a set of points equidistant from a fixed point called the centre. The distance around the entire circle boundary is called the circumference. The radius is any line segment joining its centre to any point on the circumference. radius centre The diameter is a line segment passing through the centre. Note that the diameter of a circle is twice its radius circumference www.mathssupport.org
Naming the parts of a circle A chord is any line segment that joins two points on the circle. Therefore, a diameter is an example of a chord. It is the longest possible chord. The line that touches the circumference in exactly one point is called a tangent or a tangent line. The point where the tangent touches the circle is called the point of contact or point of tangency. centre www.mathssupport.org
Tangents from an external point Draw a circle, label the centre O Choose an external point and label it P From the point P draw a tangent to the circle and label A the point of tangency. From the point P draw another tangent to the circle and label B the point of tangency. P A Measure the length AP Measure the length BP O What do you notice? B Repeat the procedure with another circle and tangents AP = BP Statement Tangents from an external point are equal in length. www.mathssupport.org
Tangents from an external point Find the length of the tangents AP and BP if the radius is 8cm, AP is tangent at A and PB is tangent at B. P is located 20 cm from the center O. Solution: As OA is the radius, A is the point of tangency and AP is a tangent, perpendicular to OA The angle OAP is 90o so, the triangle OAP is a right angled triangle P A 20 cm Using Pythagoras theorem 202 = 82 + AP2 400 = 64 + AP2 400 - 64 = AP2 8 cm O B 336 = AP2 18.3 = AP The length AP and BP is 18.3cm www.mathssupport.org
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