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How do you solve tangent problems in geometry? - Answers

You find the ( x,y) coordinates of the point where the tangent line just touches the circumference of a circle. The tangent line will have the straight line eq'n ( y = mx + c) The circle circumference will have the circle eq'n ( x^(2) + y^(2) = r^(2) The way to do it is to square up the straight line eq;b/ y^(2) = (mx + c)^(2) = ( (mx)^(2) + 2mxc + c^(2) ) Note how the RHS is squared up. Subtract to eliminate 'y' Then you have only 'x' to be solved. Since 'x' is squared 'x^(2)' , and the 'm' , 'r' and 'c' are constants, you can form a quadratic eq'n, to solve for 'x'. Once solved you substitute 'x' back into either eq'n to find 'y'.



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How do you solve tangent problems in geometry? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_solve_tangent_problems_in_geometry

You find the ( x,y) coordinates of the point where the tangent line just touches the circumference of a circle. The tangent line will have the straight line eq'n ( y = mx + c) The circle circumference will have the circle eq'n ( x^(2) + y^(2) = r^(2) The way to do it is to square up the straight line eq;b/ y^(2) = (mx + c)^(2) = ( (mx)^(2) + 2mxc + c^(2) ) Note how the RHS is squared up. Subtract to eliminate 'y' Then you have only 'x' to be solved. Since 'x' is squared 'x^(2)' , and the 'm' , 'r' and 'c' are constants, you can form a quadratic eq'n, to solve for 'x'. Once solved you substitute 'x' back into either eq'n to find 'y'.



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https://math.answers.com/math-and-arithmetic/How_do_you_solve_tangent_problems_in_geometry

How do you solve tangent problems in geometry? - Answers

You find the ( x,y) coordinates of the point where the tangent line just touches the circumference of a circle. The tangent line will have the straight line eq'n ( y = mx + c) The circle circumference will have the circle eq'n ( x^(2) + y^(2) = r^(2) The way to do it is to square up the straight line eq;b/ y^(2) = (mx + c)^(2) = ( (mx)^(2) + 2mxc + c^(2) ) Note how the RHS is squared up. Subtract to eliminate 'y' Then you have only 'x' to be solved. Since 'x' is squared 'x^(2)' , and the 'm' , 'r' and 'c' are constants, you can form a quadratic eq'n, to solve for 'x'. Once solved you substitute 'x' back into either eq'n to find 'y'.

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      You find the ( x,y) coordinates of the point where the tangent line just touches the circumference of a circle. The tangent line will have the straight line eq'n ( y = mx + c) The circle circumference will have the circle eq'n ( x^(2) + y^(2) = r^(2) The way to do it is to square up the straight line eq;b/ y^(2) = (mx + c)^(2) = ( (mx)^(2) + 2mxc + c^(2) ) Note how the RHS is squared up. Subtract to eliminate 'y' Then you have only 'x' to be solved. Since 'x' is squared 'x^(2)' , and the 'm' , 'r' and 'c' are constants, you can form a quadratic eq'n, to solve for 'x'. Once solved you substitute 'x' back into either eq'n to find 'y'.

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