+17 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
+2440 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.
