The equation of the circle with centre on the | vedclass

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Question

(A) Let the centre of the circle be $(h, 0)$ since it lies on the $x$-axis. The radius is $r = 5$.The equation of the circle is $(x - h)^2 + (y - 0)^2

Vedclass · Accepted Answer

$x^2 + y^2 + 4x - 21 = 0$

Vedclass · Answer

(A) Let the centre of the circle be $(h, 0)$ since it lies on the $x$-axis. The radius is $r = 5$.The equation of the circle is $(x - h)^2 + (y - 0)^2 = 5^2$,which simplifies to $(x - h)^2 + y^2 = 25$.Since the circle passes through the point $(2, 3)$,we substitute these coordinates into the equation:$(2 - h)^2 + 3^2 = 25$$(2 - h)^2 + 9 = 25$$(2 - h)^2 = 16$$2 - h = \pm 4$Case $1$: $2 - h = 4 \Rightarrow h = -2$. The equation is $(x + 2)^2 + y^2 = 25$ $\Rightarrow x^2 + 4x + 4 + y^2 = 25$ $\Rightarrow x^2 + y^2 + 4x - 21 = 0$.Case $2$: $2 - h = -4 \Rightarrow h = 6$. The equation is $(x - 6)^2 + y^2 = 25$ $\Rightarrow x^2 - 12x + 36 + y^2 = 25$ $\Rightarrow x^2 + y^2 - 12x + 11 = 0$.Comparing with the given options,the correct equation is $x^2 + y^2 + 4x - 21 = 0$.

The equation of the circle with centre on the $x$-axis,radius $5$ and passing through the point $(2, 3)$ is:

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