For some positive real number r , the line x+y=r is tangent to the circle x^2+y^2 = r. Find r.
How do we do this? Set equal equations together??
r^1/2 is the radius of the circle.
r is the y-intercept of the line
y = r - x.
It is tangent to the circle if
r^2 = 2*(sqrtr)^2
(from Pythagorean rule)
Therefore r^2 = 2r
r = 2
Plot the two functions
y = 2-x and
x^2 + y^2 = 2
You will see that there is tangency at
(sqrt2, sqrt2)
thanks a bunch sire
Yes set the two equations equal to each other. You then get that r = 2. I'm not going to tell you all of the work because drwls has already answered.
I am so confused by the symbols and square roots. I know what they mean, but there are so many of them!
Allow me to try to explain it without a surplus of symbols for people that don't frequent doing the math on the keyboard.
First, change the first equation to create a linear one. y=(-x)+r
The other equation is one of a circle. You should know this.
(x-h)²+(y-k)²=r²
Since x²+y²=r, the radius is √r.
That means the radius of the circle should touch the line at a point of tangency √r from the origin of the circle, 0,0. The radius which touches the point of tangency is perpendicular to y=(-x)+r.
Now you could do it the way drwls did, or you could use the formula for the length between a point and a line. Either way, you get r=2.
Well, let's see if we can solve this mathematically while keeping our sense of humor intact!
First, we'll need to find the point where the line and the circle intersect. To do this, we'll substitute y = r - x into the equation of the circle:
x^2 + (r - x)^2 = r
Expanding and simplifying, we get:
x^2 + r^2 - 2rx + x^2 = r
2x^2 - 2rx + r^2 - r = 0
Now, to find the values of x where the line and the circle intersect, the discriminant of this quadratic equation must be zero (since the line is tangent to the circle):
(2r)^2 - 4(2)(r^2 - r) = 0
Simplifying further:
4r^2 - 8r^2 + 8r = 0
-4r^2 + 8r = 0
Factor out -4r:
-4r(r - 2) = 0
This equation gives us two possible values for r: r = 0, or r = 2.
Since we're looking for a positive real number r, we can discard r = 0. Therefore, the only solution is r = 2.
So, the line x + y = r is tangent to the circle x^2 + y^2 = r when r = 2.
And there you have it! Mathematically finding the value of r can be as fun as being a clown at the circus!
To find the value of "r" for which the line x+y=r is tangent to the circle x^2+y^2=r, we need to find the point of tangency between the line and the circle.
First, let's find the equation of the line x+y=r in terms of "x" and "y". Since the equation of a line can be written in the form y = mx + b, where "m" is the slope and "b" is the y-intercept, let's rewrite the given equation as y = -x + r.
Next, let's substitute this equation into the equation of the circle x^2+y^2=r. We get x^2 + (-x + r)^2 = r. Expanding and simplifying, we have x^2 + (x^2 - 2rx + r^2) = r. Combining like terms, we get 2x^2 - 2rx + r^2 = r.
Now, we know that for a line to be tangent to a circle, the point of tangency lies on both the line and the circle. So, we need to find the value(s) of "x" that satisfy this equation.
To do this, let's solve for "x" by setting the quadratic equation equal to zero. We have 2x^2 - 2rx + r^2 - r = 0. To make the solution easier, let's simplify the equation by dividing all terms by 2: x^2 - rx + (r^2 - r/2) = 0.
Since we are looking for a point of tangency, the equation should have only one solution, which means the discriminant (b^2 - 4ac) should be equal to zero. In this case, the discriminant is r^2 - 4(r^2 - r/2) = 0. Expanding and further simplifying, we have r^2 - 4r^2 + 2r = 0.
Combine like terms, we get -3r^2 + 2r = 0. Factoring out an "r", we have r(-3r + 2) = 0. Setting each factor equal to zero, we get r = 0 and -3r + 2 = 0.
Since "r" cannot be zero (as it represents the radius of the circle), let's solve the second equation: -3r + 2 = 0. Solving for "r", we have -3r = -2, which gives r = 2/3.
Therefore, the value of "r" for which the line x+y=r is tangent to the circle x^2+y^2=r is r = 2/3.