1. **Problem Statement:** (a) Points $P(x_1,y_1)$, $Q(x_2,y_2)$, and $R(x,y)$ lie on a circle with $PQ$ as the diameter. Show the equation of the circle. (b) Find the center and vertices of the ellipse given by $3x^2 + 2y^2 - 24x + 8y + 20 = 0$. 2. **Part (a) - Equation of the Circle:** - The circle with diameter $PQ$ has the property that any point $R$ on the circle satisfies the right angle condition: $\angle PRQ = 90^\circ$. - Using the dot product of vectors $\overrightarrow{RP}$ and $\overrightarrow{RQ}$, we have: $$\overrightarrow{RP} \cdot \overrightarrow{RQ} = 0$$ - Expressing vectors: $$\overrightarrow{RP} = (x_1 - x, y_1 - y), \quad \overrightarrow{RQ} = (x_2 - x, y_2 - y)$$ - Dot product: $$ (x_1 - x)(x_2 - x) + (y_1 - y)(y_2 - y) = 0 $$ - This is the equation of the circle with $PQ$ as diameter. 3. **Part (b) - Center and Vertices of the Ellipse:** - Given ellipse equation: $$3x^2 + 2y^2 - 24x + 8y + 20 = 0$$ - Complete the square for $x$ and $y$ terms: For $x$: $$3x^2 - 24x = 3(x^2 - 8x) = 3[(x^2 - 8x + 16) - 16] = 3(x - 4)^2 - 48$$ For $y$: $$2y^2 + 8y = 2(y^2 + 4y) = 2[(y^2 + 4y + 4) - 4] = 2(y + 2)^2 - 8$$ - Substitute back: $$3(x - 4)^2 - 48 + 2(y + 2)^2 - 8 + 20 = 0$$ - Simplify constants: $$3(x - 4)^2 + 2(y + 2)^2 - 36 = 0$$ - Move constant to right side: $$3(x - 4)^2 + 2(y + 2)^2 = 36$$ - Divide both sides by 36: $$\frac{(x - 4)^2}{12} + \frac{(y + 2)^2}{18} = 1$$ - The ellipse is centered at $(4, -2)$. - Semi-major axis $a = \sqrt{18} = 3\sqrt{2}$ along $y$-axis. - Semi-minor axis $b = \sqrt{12} = 2\sqrt{3}$ along $x$-axis. - Vertices along $y$-axis: $$ (4, -2 \pm 3\sqrt{2}) $$ - Vertices along $x$-axis: $$ (4 \pm 2\sqrt{3}, -2) $$ **Final answers:** (a) Equation of circle: $$ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 $$ (b) Center: $(4, -2)$ Vertices: $(4, -2 \pm 3\sqrt{2})$ and $(4 \pm 2\sqrt{3}, -2)$