The equation of a circle with centre (5,4) an | vedclass

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(D) The centre of the circle is $(h, k) = (5, 4)$.Since the circle touches the $Y$-axis,the radius $r$ is equal to the absolute value of the $x$-coord

Vedclass · Accepted Answer

$x^2+y^2-10x-8y+16=0$

Vedclass · Answer

(D) The centre of the circle is $(h, k) = (5, 4)$.Since the circle touches the $Y$-axis,the radius $r$ is equal to the absolute value of the $x$-coordinate of the centre.Thus,$r = |5| = 5$.The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$.Substituting the values,we get $(x-5)^2 + (y-4)^2 = 5^2$.Expanding this,we have $(x^2 - 10x + 25) + (y^2 - 8y + 16) = 25$.Simplifying,$x^2 + y^2 - 10x - 8y + 41 = 25$.Therefore,$x^2 + y^2 - 10x - 8y + 16 = 0$.

The equation of a circle with centre $(5,4)$ and touching the $Y$-axis is

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