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(A) Let the centre of the circle be $(h, k)$. Since the centre lies on the line $y = x - 1$,we have $k = h - 1$,or $h - k = 1$ ... $(i)$.The radius of

Vedclass · Accepted Answer

${x^2} + {y^2} - 8x - 6y + 16 = 0$

Vedclass · Answer

(A) Let the centre of the circle be $(h, k)$. Since the centre lies on the line $y = x - 1$,we have $k = h - 1$,or $h - k = 1$ ... $(i)$.The radius of the circle is $r = 3$. The equation of the circle is $(x - h)^2 + (y - k)^2 = 3^2 = 9$.Since the circle passes through the point $(7, 3)$,we substitute these coordinates into the equation:$(7 - h)^2 + (3 - k)^2 = 9$ ... $(ii)$.Substitute $k = h - 1$ into equation $(ii)$:$(7 - h)^2 + (3 - (h - 1))^2 = 9$$(7 - h)^2 + (4 - h)^2 = 9$$(49 - 14h + h^2) + (16 - 8h + h^2) = 9$$2h^2 - 22h + 65 = 9$$2h^2 - 22h + 56 = 0$$h^2 - 11h + 28 = 0$$(h - 4)(h - 7) = 0$.Case $1$: If $h = 4$,then $k = 4 - 1 = 3$. The centre is $(4, 3)$. The equation is $(x - 4)^2 + (y - 3)^2 = 9$,which simplifies to $x^2 - 8x + 16 + y^2 - 6y + 9 = 9$,or $x^2 + y^2 - 8x - 6y + 16 = 0$.Case $2$: If $h = 7$,then $k = 7 - 1 = 6$. The centre is $(7, 6)$. The equation is $(x - 7)^2 + (y - 6)^2 = 9$,which simplifies to $x^2 - 14x + 49 + y^2 - 12y + 36 = 9$,or $x^2 + y^2 - 14x - 12y + 76 = 0$.Comparing with the given options,the equation $x^2 + y^2 - 8x - 6y + 16 = 0$ is present.

$A$ circle has a radius of $3$ units and its centre lies on the line $y = x - 1$. If the circle passes through the point $(7, 3)$,what is its equation?

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