A circle has radius 3 units and its center lies on the line $y = x - 1$ . Find the equation of the circle of it passes through $(7, 3)$

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Solution

If (h, k) be the centre then $k = h - 1$ as it lies on $y = x - 1$ Also r = 3
$∴,$ Equation is $(x - h)^{2} + (y - h + 1)^{2} = 9$
It passes through the point (7,3)
$∴ h^{2} - 11 h + 28 = 0$
$∴ h = 4, 7$

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A circle has radius 3 units and its center lies on the line y=x−1 . Find the equation of the circle of it passes through (7,3)

If (h, k) be the centre then k=h−1 as it lies on y=x−1 Also r = 3∴, Equation is(x−h)2+(y−h+1)2=9It passes through the point (7,3) ∴h2−11h+28=0∴h=4,7

A circle has radius 3 units and its center lies on the line $y = x - 1$ . Find the equation of the circle of it passes through $(7, 3)$

If (h, k) be the centre then $k = h - 1$ as it lies on $y = x - 1$ Also r = 3
$∴,$ Equation is $(x - h)^{2} + (y - h + 1)^{2} = 9$
It passes through the point (7,3)
$∴ h^{2} - 11 h + 28 = 0$
$∴ h = 4, 7$