Tangents and Circles | MathHelp.com - By MathHelp.com
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| 00:0-1 | in the diagram shown line S . R . Is | |
| 00:03 | tangent to circle . Oh if O T equals nine | |
| 00:08 | and T . S . Equals six . Find S | |
| 00:11 | . R . Our first step in this problem is | |
| 00:15 | to label the diagram , We know that O . | |
| 00:18 | T . equals nine . Yeah & T . S | |
| 00:24 | Equals six . Therefore Os yeah yeah equals nine plus | |
| 00:34 | six or 15 . And since we're given that O | |
| 00:41 | . T equals nine , we know that O . | |
| 00:43 | R . Must also equal nine . Since O . | |
| 00:48 | T . And O . R are both radi I | |
| 00:51 | of circle . Oh and already I of a circle | |
| 00:54 | are congruent . Now we're asked to find S . | |
| 00:58 | R . So let's use X . To represent the | |
| 01:02 | length of segment S . R . To find the | |
| 01:05 | value of X . Notice that were given that line | |
| 01:09 | S . R . Is tangent to circle . Oh | |
| 01:13 | and remember that if a line is tangent to a | |
| 01:16 | circle then the line is perpendicular to the radius of | |
| 01:20 | the circle drawn to the point of tangent . C | |
| 01:24 | . So we know that line S . R . | |
| 01:26 | Is perpendicular to radius . O . R wow . | |
| 01:31 | This means that triangle S . R . O . | |
| 01:35 | Is a right triangle . So to find the value | |
| 01:38 | of X , we can use the pythagorean theorem to | |
| 01:42 | set up the equation X squared plus nine squared equals | |
| 01:48 | 15 squared yeah . Uh huh . Which simplifies to | |
| 01:58 | X squared plus 81 Equals 225 . And subtracting 81 | |
| 02:08 | from both sides mm gives us X squared Equals 144 | |
| 02:19 | and square rooting both sides , We find that x | |
| 02:23 | equals 12 , So the length of segment S . | |
| 02:31 | R . is 12 . Yeah , wow . |
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