5.02c - Circle and Tangents | Lecture Note

University:
High School
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High School Physics
Academic year:

2024
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5.02c Circles and Tangency Tangents to Circles • A tangent touches the circle at only one point (the point of tangency) and is perpendicular to the radial arm it touches. Steps to Find the Rule 1. Sketch the circle as it is described. 2. Find the slope of the radial arm. 3. The slope of the tangent line is the negative reciprocal. 4. Find the equation using the slope of the tangent line and the point of tangency. Steps in Action A point P(-2,3) lies on the circle with equation (𝑥 + 3)2 +(𝑦 − 5)2 = 5. Find the equation of the tangent line through P. 1. Draw the circle as it’s described 2. Slope of radius line 𝑎= ∆𝑦 ∆𝑥 = 𝑦2 −𝑦1 𝑥2 −𝑥1 C (-3, 5) = P (-2, 3) 5 −(3) (−3)−(−2) 3. Slope of the tangent line 𝑎= 1 2 = 2 − 1 Steps in Action A point P(-2,3) lies on the circle with equation (𝑥 + 3)2 +(𝑦 − 5)2 = 5. Find the equation of the tangent line through P. 1. Draw the circle as it’s described 2. Slope of radius line 𝑎=− C (-3, 5) 2 1 3. Slope of the tangent line P (-2, 3) 4. Find equation 1 −2,3 → 𝑦 = 𝑥 + 𝑏 2 𝑏=4 𝑎= 1 2 1 𝑦 = 𝑥+4 2 𝒙 − 𝟐𝒚 + 𝟖 = 𝟎 Finding Length of Tangent to any point Find the length of a tangent through point Q(2,8) to the circle (𝑥 + 1)2 +(𝑦 − 1)2 = 12 1. Draw the circle as it’s described 2. Consider right Triangle 4. Find QT Pythagorean Theorem 𝑎2 + 𝑏2 = 𝑐 2 3. Find d(O,Q) 𝑑 𝑂, 𝑄 = ∆𝑥 2 + ∆𝑦 2 (2 − −1 )2 +(8 − 1)2 (3)2 +(7)2 58 𝑏2 = 2 58 − 𝑏2 = 46 𝑏 = 46 12 2 ∴ 𝑃𝑜𝑖𝑛𝑡 𝑇 𝑖𝑠 46 𝑢𝑛𝑖𝑡𝑠 𝑓𝑟𝑜𝑚 𝑃𝑜𝑖𝑛𝑡 𝑄 Finding Intersections Point(s) 1. Substitute the line equation (with one variable isolated) into the circle equation 2. Solve the quadratic equation: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 use factoring or quadratic formula 3. Possibilities – check the discriminant (𝑏 2 − 4𝑎𝑐) ∆< 0 (No intersection) ∆= 0 (1 intercept) ∆> 0 (2 intercepts) Finding Intersections Point(s) Find intersection(s) of (𝑥 − 2)2 +(𝑦 + 1)2 = 16 and 𝑦 = 2𝑥 − 1 Solve the System (𝑥 − 2)2 +(𝟐𝒙 − 𝟏 + 1)2 = 16 𝑥 2 − 4𝑥 + 4 + 4𝑥 2 = 16 5𝑥 2 − 4𝑥 − 12 = 0 Discriminant 𝑏2 − 4𝑎𝑐 (−4)2 −4(5)(−12) 256 ∴ secant Quadratic Equation −𝑏 ± 𝑏 2 − 4𝑎𝑐 −(−4) ± (−4)2 −4(5)(−12) 𝑥= → 2𝑎 2(5) 𝑥1 = 6 − 5 𝑥2 = 2 𝑦1 = 17 − 5 𝑦2 = 3 Finding Intersections Point(s) Find intersection(s) of (𝑥 + 4)2 +(𝑦 − 5)2 = 4 and 0 = 2𝑥 − 𝑦 + 3 Solve the System (𝑥 + 4)2 +(𝟐𝒙 + 𝟑 − 5)2 = 4 𝑥 2 + 8𝑥 + 16 + 4𝑥 2 − 8𝑥 + 4 = 4 5𝑥 2 + 16 = 0 Discriminant 𝑏 2 − 4𝑎𝑐 02 − 4(5)(16) = −320 < 0 ∴ 𝑁𝑜 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑠 Homework #14, #15 Homework • #30, #32