The equation of the circle which touches both | vedclass

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(B) The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Given the centre is $(x_1, y_1)$ and the ci

Vedclass · Accepted Answer

${x^2} + {y^2} - 2{x_1}(x + y) + x_1^2 = 0$

Vedclass · Answer

(B) The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Given the centre is $(x_1, y_1)$ and the circle touches both axes,the radius $r$ must satisfy $r = |x_1| = |y_1|$.Thus,$x_1 = y_1 = r$ (assuming the circle is in the first quadrant,$x_1^2 = y_1^2 = r^2$).Substituting these into the equation: $(x - x_1)^2 + (y - x_1)^2 = x_1^2$.Expanding the terms: $(x^2 - 2x x_1 + x_1^2) + (y^2 - 2y x_1 + x_1^2) = x_1^2$.Simplifying: $x^2 + y^2 - 2x_1(x + y) + x_1^2 = 0$.

The equation of the circle which touches both axes and whose centre is $({x_1}, {y_1})$ is

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