D12 (p. 16)

[BenCurtis]

Find the equation of the line that is tangent to the circle x^2 + y^2 = 25 at (5/2(√(2),5/2(√(2)). I started with this equation: (5/2(√(2))x + (5/2(√(2))y - 25 = 0. I moved the 25 to the right side of the equation (making it positive) then factored out the 5/2 and then the √(2). This left me with x + y = 10/√(2). The correct answer is x + y = 5√(2). I need help.

 

[MarkFL]

If you draw a diagram, you see that the tangent line is perpendicular to the line \(\displaystyle y=x\), and you are given the point through which it passes, so what does this tell you?

 

[BenCurtis]

Well, this is what I know from my diagram. If the tangent line is perpendicular to y=x, then the tangent line is -y=x. Also, we know that one point on the line is (5/2(√(2),5/2(√(2)). I know that the equation for a line is y=mx + b. Combining the equation for a line w/ -y=x gives you mx - y=-b. We also know that m=-1 so our equation is -x - y=-b or x + y=b. If we solve for b, the y=intercept, y becomes equal to 0. From the first triangle in my sketch, I see that the distance from the origin to x is 5/2(√(2), as given. The second triangle in my sketch (starting from x=5/2(√(2) to where the tan line crosses the x-axis) is 2 * 5/2(√(2) or 5(√(2).

 

[HallsofIvy]

Well, this is what I know from my diagram. If the tangent line is perpendicular to y=x, then the tangent line is -y=x.

No, thats not true. -y= x passes through the origin which is the center of the circle so is not tangent. Perhaps you meant that the tangent line is parallel to -y= x. Or, just that, because the given line ius y= x, the slope of a line perpendicular to is -1.

Also, we know that one point on the line is (5/2(√(2),5/2(√(2)). I know that the equation for a line is y=mx + b. Combining the equation for a line w/ -y=x

I would think that just "y= -x+ b" would be simpler.

gives you mx - y=-b. We also know that m=-1 so our equation is -x - y=-b or x + y=b. If we solve for b, the y=intercept, y becomes equal to 0. The y intercept is, by definition, the y value when x is equal to 0. If you mean that, for this particular problem, y is also 0, that is not true.

From the first triangle in my sketch, I see that the distance from the origin to x is 5/2(√(2), as given. The second triangle in my sketch (starting from x=5/2(√(2) to where the tan line crosses the x-axis) is 2 * 5/2(√(2) or 5(√(2).

I don't know what "triangles" you are referring to but you wrote the point of tangency as \(\displaystyle (5/2(\sqrt{2}), 5/2(\sqrt{2}))\) that is ambiguous as to whether you meant \(\displaystyle ((5/2)\sqrt{2}, (5/2)\sqrt{2})\) or \(\displaystyle (5/(2\sqrt{2}), 5/(2\sqrt{2}))\) but only the first satisfies \(\displaystyle x^2+ y^2= 5\)
And, while writing a lot (often the same thing repeatedly) you never did write the answer. (And you are missing right parentheses in 2 * 5/2(√(2) and 5(√(2).) If \(\displaystyle y= -x+ b\) and \(\displaystyle y= (5/2)\sqrt{2}\) when \(\displaystyle x= (5/2)\sqrt{2}\), then \(\displaystyle (5/2)\sqrt{2}= -(5/2)\sqrt{2}+ b\) so that \(\displaystyle b= 5\sqrt{2}\) so your answer should be \(\displaystyle y= -x+ 5\sqrt{2}\).

 

[BenCurtis]

HallsofIvy, yes, you're right that I meant to say the line perpendicular to y = x is has a slope of -1. And you're right, the y-intercept is where the line crosses the y-axis: x is 0, but the value of y can vary. As to what I meant, w/ regards to my coordinates it's suppose to be ((5/2)*√(2),(5/2)*√(2)). I'll try to be more careful about my parentheses. Thank you for explaining how you came up w/ x + y = 5(√(2). You used the equation of a line having a slope of -1 and substituted (5/2)*√(2) for y and x, as given.