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i made a mistake on my question from before: i meant 5 not 25! sorry!! thanks again!

 

prove algebraically that the straight line with the equation x = 2y + 5 is a tangent to the circle with equation x^2 + y^2 = 5

 

 
andrxxa  Sep 27, 2018
edited by Guest  Sep 27, 2018
edited by andrxxa  Sep 27, 2018
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A line is tangent to a circle if it intersects that circle at exactly one point. That's all we need to prove. 

 

\(\boxed{1}\hspace{1mm}x=2y+5\\ \boxed{2}\hspace{1mm}x^2+y^2=5\) We can use the method of substitution to find the point of intersection.
\((2y+5)^2+y^2=5\) We have to expand the square of a binomial in order to make some progress here. 
\(4y^2+20y+25+y^2=5\) Combine the like terms and bring all terms to one side of the equation.
\(5y^2+20y+20=0\) Factor out the greatest common factor from every term, 5. 
\(5(y^2+4y+4)=0\) The trinomial contained in the parentheses can be written as a square of a binomial. 
\(5(y+2)^2=0\) Divide by 5 from both sides of the equation.
\((y+2)^2=0\) Take the square root of both sides.
\(\sqrt{(y+2)^2}=\sqrt{0}\) Write the left-hand side as the absolute value of the argument.
\(|y+2|=0\) Solve for y. 
\(y+2=0\)  
\(y=-2\) Now, solve for x. Since we are proving that this line is tangent, we should have foreseen that there would only be one possible y-value. Now, we must solve for the x-coordinate.
   

 

\(\boxed{1}\hspace{1mm}x=2y+5; y=-2\) Of course, you may substitute into either equation, but it is probably desirable to substitute the known y-value into this equation because it is most likely easier.
\(x=-2*2+5\) Simplify.
\(x=1\)  
   

 

Therefore, the intersection point of the line and circle is located at only \((1,-2)\). Since there is only one intersection point, the line is tangent to the circle.

TheXSquaredFactor  Sep 27, 2018

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