If a circle touches both axes and lies below | vedclass

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(A) Since the circle touches both axes in the first quadrant,its center is $(r, r)$ and radius is $r$,where $r > 0$.The equation of the circle is $(x

Vedclass · Accepted Answer

$4x^2 + 4y^2 - 4x - 4y + 1 = 0$

Vedclass · Answer

(A) Since the circle touches both axes in the first quadrant,its center is $(r, r)$ and radius is $r$,where $r > 0$.The equation of the circle is $(x - r)^2 + (y - r)^2 = r^2$,which simplifies to $x^2 + y^2 - 2rx - 2ry + r^2 = 0$.The distance from the center $(r, r)$ to the line $4x + 3y - 6 = 0$ must be greater than the radius $r$ for the circle to lie below the line.The perpendicular distance $d = \frac{|4r + 3r - 6|}{\sqrt{4^2 + 3^2}} = \frac{|7r - 6|}{5}$.For the circle to be below the line,we require $d > r$ or specific conditions. However,if the circle is tangent to the line,$d = r$.$\frac{|7r - 6|}{5} = r \implies |7r - 6| = 5r$.Case $1$: $7r - 6 = 5r \implies 2r = 6 \implies r = 3$.Case $2$: $7r - 6 = -5r \implies 12r = 6 \implies r = 0.5$.For $r = 0.5$,the circle is $(x - 0.5)^2 + (y - 0.5)^2 = 0.25$,which is $x^2 + y^2 - x - y + 0.25 = 0$,or $4x^2 + 4y^2 - 4x - 4y + 1 = 0$.

If a circle touches both axes and lies below the line $4x + 3y = 6$ in the first quadrant,then the equation of the circle is:

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