The equation of a circle is $(x-3)^2+y^2=18$. The line with equation $y=m x+c$ passes through the point $(0,-9)$ and is a tangent to the circle.
Find the two possible values of $m$ and, for each value of $m$, find the coordinates of the point at which the tangent touches the circle.
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Answer (2)
The two possible values for the slope m of the tangent line are 1 and − 7 . The points at which the tangent touches the circle are ( 6 , − 3 ) for m = 1 and ( − 5 6 , − 5 3 ) for m = − 7 .
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Find the value of c by substituting the point ( 0 , − 9 ) into the equation y = m x + c , resulting in c = − 9 .
Use the point-to-line distance formula to set up an equation relating the distance from the circle's center to the tangent line with the radius, and solve for m , obtaining m = 1 and m = − 7 .
For each value of m , substitute the tangent line equation into the circle equation and solve for x to find the x-coordinate of the tangency point. Then, substitute x back into the tangent line equation to find the y-coordinate.
State the two possible values of m and their corresponding points of tangency: m = 1 with point ( 6 , − 3 ) , and m = − 7 with point ( − 5 6 , − 5 3 ) . The final answer is: m = 1 , ( 6 , − 3 ) ; m = − 7 , ( − 5 6 , − 5 3 )
Explanation
Problem Analysis The equation of the circle is ( x − 3 ) 2 + y 2 = 18 . The center of the circle is ( 3 , 0 ) and the radius is 18 = 3 2 . The equation of the tangent line is y = m x + c . The tangent line passes through the point ( 0 , − 9 ) . We need to find the two possible values of m and, for each value of m , find the coordinates of the point at which the tangent touches the circle.
Finding c Since the line y = m x + c passes through ( 0 , − 9 ) , we can substitute x = 0 and y = − 9 into the equation to find c . Thus, − 9 = m ( 0 ) + c , which gives c = − 9 . Therefore, the equation of the tangent line is y = m x − 9 .
Using Point-to-Line Distance The distance from the center of the circle ( 3 , 0 ) to the tangent line y = m x − 9 is equal to the radius of the circle 3 2 . We can use the point-to-line distance formula: A 2 + B 2 ∣ A x 0 + B y 0 + C ∣ . Rewrite the line equation as m x − y − 9 = 0 .
Applying the Formula Applying the point-to-line distance formula with ( x 0 , y 0 ) = ( 3 , 0 ) , A = m , B = − 1 , and C = − 9 , the distance is m 2 + ( − 1 ) 2 ∣3 m − 0 − 9∣ = m 2 + 1 ∣3 m − 9∣ .
Setting up the Equation Setting the distance equal to the radius, we have m 2 + 1 ∣3 m − 9∣ = 3 2 . Squaring both sides, we get m 2 + 1 ( 3 m − 9 ) 2 = 18 , which simplifies to ( 3 m − 9 ) 2 = 18 ( m 2 + 1 ) . Expanding, we have 9 m 2 − 54 m + 81 = 18 m 2 + 18 . Rearranging, we get 9 m 2 + 54 m − 63 = 0 . Dividing by 9, we have m 2 + 6 m − 7 = 0 .
Solving for m Factoring the quadratic equation, we get ( m + 7 ) ( m − 1 ) = 0 . Thus, the possible values of m are m = 1 and m = − 7 .
Finding Tangency Point for m=1 For m = 1 , the equation of the tangent line is y = x − 9 . To find the point of tangency, we solve the system of equations formed by the circle ( x − 3 ) 2 + y 2 = 18 and the tangent line y = x − 9 . Substituting y = x − 9 into the circle equation, we get ( x − 3 ) 2 + ( x − 9 ) 2 = 18 . Expanding, we have x 2 − 6 x + 9 + x 2 − 18 x + 81 = 18 , which simplifies to 2 x 2 − 24 x + 72 = 0 . Dividing by 2, we get x 2 − 12 x + 36 = 0 . Factoring, we have ( x − 6 ) 2 = 0 , so x = 6 . Substituting x = 6 into y = x − 9 , we get y = 6 − 9 = − 3 . Thus, the point of tangency is ( 6 , − 3 ) .
Finding Tangency Point for m=-7 For m = − 7 , the equation of the tangent line is y = − 7 x − 9 . Substituting y = − 7 x − 9 into the circle equation ( x − 3 ) 2 + y 2 = 18 , we get ( x − 3 ) 2 + ( − 7 x − 9 ) 2 = 18 . Expanding, we have x 2 − 6 x + 9 + 49 x 2 + 126 x + 81 = 18 , which simplifies to 50 x 2 + 120 x + 72 = 0 . Dividing by 2, we get 25 x 2 + 60 x + 36 = 0 . Factoring, we have ( 5 x + 6 ) 2 = 0 , so x = − 5 6 . Substituting x = − 5 6 into y = − 7 x − 9 , we get y = − 7 ( − 5 6 ) − 9 = 5 42 − 5 45 = − 5 3 . Thus, the point of tangency is ( − 5 6 , − 5 3 ) .
Final Answer The two possible values of m are 1 and − 7 . The point of tangency for m = 1 is ( 6 , − 3 ) , and the point of tangency for m = − 7 is ( − 5 6 , − 5 3 ) .
Examples
Understanding tangent lines and circles is crucial in various fields. For instance, in engineering, designing curved roads or circular gears requires precise calculations involving tangents to ensure smooth transitions and optimal performance. Similarly, in computer graphics, determining tangent lines is essential for rendering realistic reflections and creating smooth curves in 3D models. This problem showcases how geometric principles are applied in real-world applications to solve practical problems.
