The vertices of Delta ABC are A(a, 6), B(5, 1 | vedclass

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(D) The circumcentre $P(2, 3)$ is equidistant from all vertices of the triangle. Therefore, $PA = PB = PC$.First, calculate the distance $PB$:$PB^2 =

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$a = 4$

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(D) The circumcentre $P(2, 3)$ is equidistant from all vertices of the triangle. Therefore, $PA = PB = PC$.First, calculate the distance $PB$:$PB^2 = (5 - 2)^2 + (1 - 3)^2 = 3^2 + (-2)^2 = 9 + 4 = 13$.Since $PA^2 = PB^2$, we have:$(a - 2)^2 + (6 - 3)^2 = 13$$(a - 2)^2 + 3^2 = 13$$(a - 2)^2 + 9 = 13$$(a - 2)^2 = 4$$a - 2 = \pm 2$This gives two possible values for $a$: $a = 2 + 2 = 4$ or $a = 2 - 2 = 0$.If $a = 4$, the coordinates of $A$ are $(4, 6)$, which is the same as the coordinates of $C(4, 6)$. Since $A$ and $C$ would coincide, a triangle cannot be formed.Therefore, the only valid value is $a = 0$.

The vertices of $\Delta ABC$ are $A(a, 6)$, $B(5, 1)$, and $C(4, 6)$. If the circumcentre of the triangle is $P(2, 3)$, find the value of $a$.

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