SAT ACT. What is the radius of the circle?

SAT ACT. As shown in the figure, the circle is tangent to the x-axis and y-axis. The perpendicular length from point Q on the circle to the x-axis is 2. If the x coordinate of C is 9, What is the radius of the circle? #sat #act #digitalsat#Trigonometric functions #math#sine#cosine#tangent

[@MrPaulc222]

The first few lines, although not wrong, make a simplish puzzle far more difficult than necessary. I leave them in anyway to show how NOT to do it :)
I haven't watched yet, but I might form a square with the tangent Q as its top left point. You can then make a right triangle with lines from P to the bottom right of the square (r + 2*sqrt(2), p and the x axis (r), and the x-axis tangent to the bottom right of the square (11-r).
Equation generated is r^2 + (11 - r}^2 = (r + 2*sqrt(2))^2
Expand: r^2 + 121 - 22r + r^2 = r^2 + 4*sqrt(2)r + 8
Reduce: r^2 - 22r + 121 = 8 + 4*sqrt(2)r
Rearrange: r^2 - 22r + 113 = 4*sqrt(2)r
r^2 - (22-4*sqrt(2))r + 113 = 0 looks like a very unwieldy quadratic, and I'm thinking that there must be a simpler way.
Yes, of course there is. Doh! :)
Draw a vertical line down from P to 2 above the x axis.
The right triangle would now be (r-2), (9-r), and r.
(r-2)^2 + (9-r)^2 = r^2
Expand: r^2 - 4r + 4 + 81 - 18r + r^2 = r^2
2r^2 - 22r + 85 = r^2
r^2 - 22r + 85 = 0. That quadratic looks more manageable.
(22+or-sqrt(484 - 340))/2 = r
(22+or-12)/2 = r
r = 17 or 5.
It clearly can't be 17, so must be 5.

[@brendanward2991]

I did it slightly differently. The centre of the circle is (r,r), so the equation of the circle in (x-r)^2 + (y-r)^2 = r^2. The point (9,2) is on the circle so its coordinates satisfy the equation. Solve for r.

[@Okkk517]

I think 17 is a valid solution if we allow point C to be on the left of the center point P.

[@filipo4114]

that's a cool way to use an AI

[@archangecamilien1879]

Impossible to determine, I think, with the information in the thumbnail, lol...I tend to go by whatever I see in the thumbnails for these math problems without watching the videos...I mean, lol, you can take a circle of any size (as long as the radius is larger than 2) and draw a length 2 line perpendicular to the line on the bottom that touches it at whatever point is exactly 2 above the "bottom", lol...I mean, with only that length of 2 and nothing else, this problem is impossible to solve...at first I thought that "q" at the bottom was 9, measuring something, lol...