The equation of the circle in the first quadr | vedclass

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Question

(B) The circle is in the first quadrant and touches both axes at a distance of $5$ from the origin.Therefore,the center of the circle is $(h, k) = (5,

Vedclass · Accepted Answer

$x^2 + y^2 - 10x - 10y + 25 = 0$

Vedclass · Answer

(B) The circle is in the first quadrant and touches both axes at a distance of $5$ from the origin.Therefore,the center of the circle is $(h, k) = (5, 5)$ and the radius $r = 5$.The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 5)^2 + (y - 5)^2 = 5^2$.Expanding this,we have $(x^2 - 10x + 25) + (y^2 - 10y + 25) = 25$.Simplifying,we get $x^2 + y^2 - 10x - 10y + 25 = 0$.

The equation of the circle in the first quadrant which touches each axis at a distance $5$ from the origin is

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