1. **Problem 13:** Find the possible values of $a$ and $b$ such that the curves $$x^2 + y^2 = 10$$ and $$x^2 + y^2 + ay = b$$ intersect at right angles where $x=3$. 2. **Step 1: Understand the curves** - The first curve is a circle centered at the origin with radius $\sqrt{10}$. - The second curve is also a circle but shifted vertically depending on $a$ and $b$. 3. **Step 2: Find $y$ at $x=3$ for the first curve** $$3^2 + y^2 = 10 \implies 9 + y^2 = 10 \implies y^2 = 1 \implies y = \pm 1$$ 4. **Step 3: Use the point $(3,y)$ on the second curve** Substitute $x=3$ and $y=\pm 1$ into $$x^2 + y^2 + ay = b$$: $$9 + 1 + a(\pm 1) = b \implies 10 \pm a = b$$ 5. **Step 4: Find derivatives to get slopes of tangents** - For the first curve: $$x^2 + y^2 = 10$$ differentiate implicitly: $$2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$$ - For the second curve: $$x^2 + y^2 + ay = b$$ differentiate implicitly: $$2x + 2y \frac{dy}{dx} + a \frac{dy}{dx} = 0 \implies (2y + a) \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{2x}{2y + a}$$ 6. **Step 5: Condition for right angle intersection** The tangents are perpendicular if the product of their slopes is $-1$: $$\left(-\frac{x}{y}\right) \times \left(-\frac{2x}{2y + a}\right) = -1$$ Simplify: $$\frac{2x^2}{y(2y + a)} = -1$$ 7. **Step 6: Substitute $x=3$ and $y=\pm 1$** For $y=1$: $$\frac{2 \times 9}{1(2 \times 1 + a)} = -1 \implies \frac{18}{2 + a} = -1 \implies 18 = -2 - a \implies a = -20$$ For $y=-1$: $$\frac{18}{-1(2 \times (-1) + a)} = -1 \implies \frac{18}{-1(-2 + a)} = -1 \implies \frac{18}{2 - a} = -1 \implies 18 = -2 + a \implies a = 20$$ 8. **Step 7: Find corresponding $b$ values** Recall from step 4: - For $y=1$ and $a=-20$: $$b = 10 + a = 10 - 20 = -10$$ - For $y=-1$ and $a=20$: $$b = 10 - a = 10 - 20 = -10$$ **Final answer:** $$\boxed{(a,b) = (-20,-10) \text{ or } (20,-10)}$$