+0

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

0
167
1
+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

Sep 27, 2018
edited by Guest  Sep 27, 2018
edited by andrxxa  Sep 27, 2018

#1
+2340
+2

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.

Sep 27, 2018