Question

The coordinates of circumcenter of $\triangle$PQR, is

• A. (5, 3)
• B. (5, 2)
• C. (5, 4)
• D. (5, 6)

Answer 1

Answer: (B)

(5, 2)

Answer 2

The given curve is $y^2 - 8x - 4y + 28 = 0$. Completing the square for y: $(y^2 - 4y + 4) - 4 - 8x + 28 = 0$ $(y - 2)^2 - 8x + 24 = 0$ $(y - 2)^2 = 8(x - 3)$

This is a parabola with vertex $V(h, k) = (3, 2)$. Comparing with $(y-k)^2 = 4a(x-h)$, we have $4a = 8$, so $a = 2$. The focus is $F(h+a, k) = (3+2, 2) = (5, 2)$. The axis of symmetry is $y = k$, which is $y = 2$.

The points P(5, 6) and Q(5, -2) lie on the parabola and form the latus rectum.

The normals at P and Q meet at R. The slope of the tangent at a point $(x_1, y_1)$ on $(y-k)^2 = 4a(x-h)$ is $m_{tan} = \frac{2a}{y_1-k}$. At P(5, 6): $m_{tan, P} = \frac{2(2)}{6-2} = 1$. The slope of the normal at P is $m_{norm, P} = -1$. The equation of the normal at P(5, 6) is $y - 6 = -1(x - 5) \implies y = -x + 11$.

At Q(5, -2): $m_{tan, Q} = \frac{2(2)}{-2-2} = -1$. The slope of the normal at Q is $m_{norm, Q} = 1$. The equation of the normal at Q(5, -2) is $y - (-2) = 1(x - 5) \implies y = x - 7$.

To find R, we intersect the normals: $y = -x + 11$ $y = x - 7$ Adding the equations: $2y = 4 \implies y = 2$. Substituting $y=2$ into $y = x-7$: $2 = x-7 \implies x = 9$. So, $R = (9, 2)$.

The vertices of $\triangle$PQR are P(5, 6), Q(5, -2), R(9, 2). The circumcenter is the intersection of the perpendicular bisectors of the sides. Side PQ is a vertical line segment ($x=5$). Its midpoint is $(5, 2)$. The perpendicular bisector of PQ is the horizontal line $y = 2$. Side QR has midpoint $(7, 0)$. The slope of QR is $1$. The slope of the perpendicular bisector of QR is $-1$. The equation is $y - 0 = -1(x - 7) \implies y = -x + 7$. The circumcenter is the intersection of $y=2$ and $y=-x+7$. $2 = -x + 7 \implies x = 5$. The circumcenter is (5, 2).