A circle has its centre in the first quadrant | vedclass

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(D) Let the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(2,3)$,we have $(2-h)^2 + (3-k)^2 = r^2$. The intercept on the line $x=2$ is

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$x^2+y^2-8x-9y+30=0$

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(D) Let the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(2,3)$,we have $(2-h)^2 + (3-k)^2 = r^2$. The intercept on the line $x=2$ is $2\sqrt{r^2 - (2-h)^2} = 3$. Substituting $r^2 - (2-h)^2 = (3-k)^2$,we get $2\sqrt{(3-k)^2} = 3$,so $|3-k| = 1.5$,which means $k = 4.5$ or $k = 1.5$. The intercept on the line $y=3$ is $2\sqrt{r^2 - (3-k)^2} = 4$. Substituting $r^2 - (3-k)^2 = (2-h)^2$,we get $2\sqrt{(2-h)^2} = 4$,so $|2-h| = 2$,which means $h = 4$ or $h = 0$. Since the centre $(h,k)$ is in the first quadrant,$h, k > 0$. Testing the case $(h,k) = (4, 4.5)$: $r^2 = (2-4)^2 + (3-4.5)^2 = 4 + 2.25 = 6.25$. The equation is $(x-4)^2 + (y-4.5)^2 = 6.25$,which simplifies to $x^2 - 8x + 16 + y^2 - 9y + 20.25 = 6.25$,or $x^2 + y^2 - 8x - 9y + 30 = 0$.

$A$ circle has its centre in the first quadrant and passes through $(2,3)$. If this circle makes intercepts of length $3$ and $4$ respectively on $x=2$ and $y=3$,its equation is

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