SOLUTION: Find the equation of the circle with radius 6 that is tangent to both branches of the graph of y=|x|.
Algebra.Com
Question 246624: Find the equation of the circle with radius 6 that is tangent to both branches of the graph of y=|x|.
Answer by dabanfield(803) (Show Source): You can put this solution on YOUR website!
Let's say the center of the circle is C. Draw pependicular lines from C to the two tangent lines created by y = |x|. Let's say the two lines intersect the tangent line at points A and B respectively. Let's say the Origin is point O with coordinates (0,0).
The figure CAOB is a square with sides of length 6. Triangle CAO is a right isosceles triangle with sides of length 6. So the hypotenuse CO of CAO is sqrt(36+36) = sqrt(72) = 6*sqrt(2).
The cooredinates of C then are (0,6*sqrt(2)).
So we have a circle with radius 6 and center C at (0,6sqrt(2).
Plug these values into the equation of the standard form of a circle:
(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius of the circle.
RELATED QUESTIONS
Find an equation of the circle of radius 4 that is tangent to both branches of the graph... (answered by scott8148)
Find an equation of the circle of radius 4 that is tangent to both branches of the graph... (answered by scott8148)
Find the equation of the circle that is tangent to both x and y axes and the radius at... (answered by dkppathak)
Please find an equation of the circle that satisfies the given condition. Circle lies in... (answered by Alan3354)
Find an equation of the circle that satisfies the given conditions.
Circle lies in the... (answered by Alan3354)
Find the equation of a circle that is tangent the line x = -8 and is tangent to both... (answered by stanbon)
Find equation of circle with center (1,4) that is tangent to the line... (answered by Edwin McCravy)
what is the equation of a circle having radius of 5 and tangent to both... (answered by jim_thompson5910,Alan3354)
Please help me with this! Find the equation of a circle with it's center on line x+y=4,... (answered by fractalier)