(x-6)^{2 }+(y-3)^{2} = 25

the equation above is a circle. If the circle is translated downward a usits such that the circle is tangent tot he x-axis, the equation becomes (x-6)^{2 }+(y-3+a)^{2} = 25. What is the value of a.

To make the x axis a tangent of a do we not need to make the edge of the circle touch the x-axis? That is changing the number with the y component. the radius id the root of the number on the other side of the equations. so the radius is 5. And to get the circles edge to tought the x axis it needs to be 5 places away from it. So wouldnt a = -2?

I typed that in and it said that it was the wrong answer? why?

BLANK Aug 6, 2020

#1**+1 **

By entering -2, you made the y-component to be (y - 5)^{2} which would make the circle tangent to the x-axis --

and the circle will be above the x-axis (center at (6,5) -- however, you were instructed to translate the circle

downwards -- by making a = 5, you translated the circle upwards.

You will need the y-component to be (y + 5)^{2} -- which will make the circle tangent to the x-axis -- it will

be tangent below the x-axis.

geno3141 Aug 6, 2020