If (3,1) and (-2,4) are points on a circle S | vedclass

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(A) Let the centre of the circle be $C(\alpha, \beta)$. Since the centre lies on the line $x-y+1=0$,we have $\beta = \alpha+1$. Thus,$C = (\alpha, \al

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$x=-1+\sqrt{17} \cos 	heta, y=\sqrt{17} \sin 	heta$

Vedclass · Answer

(A) Let the centre of the circle be $C(\alpha, \beta)$. Since the centre lies on the line $x-y+1=0$,we have $\beta = \alpha+1$. Thus,$C = (\alpha, \alpha+1)$.Given points $P(3,1)$ and $Q(-2,4)$ lie on the circle,so $CP^2 = CQ^2$.$(\alpha-3)^2 + (\alpha+1-1)^2 = (\alpha+2)^2 + (\alpha+1-4)^2$$(\alpha-3)^2 + \alpha^2 = (\alpha+2)^2 + (\alpha-3)^2$$\alpha^2 = (\alpha+2)^2$$\alpha^2 = \alpha^2 + 4\alpha + 4$$4\alpha = -4 \Rightarrow \alpha = -1$.So,the centre is $C(-1, 0)$.The radius $r$ is the distance $CP = \sqrt{(-1-3)^2 + (0-1)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16+1} = \sqrt{17}$.The parametric equations of a circle with centre $(h, k)$ and radius $r$ are $x = h + r \cos 	heta$ and $y = k + r \sin 	heta$.Substituting $h=-1, k=0, r=\sqrt{17}$,we get $x = -1 + \sqrt{17} \cos 	heta$ and $y = \sqrt{17} \sin 	heta$.

If $(3,1)$ and $(-2,4)$ are points on a circle $S$ whose centre lies on the line $x-y+1=0$,then the parametric equations of $S$ are:

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