 Suggested languages for you:

Europe

Problem 5

# If $$f(x+y)=f(x) f(y)$$ for all $$x$$ and $$y, f(1)=2$$ and $\alpha_{n}=f(n), n \in N$$then the equation of the circle having$$\left(\alpha_{1}, \alpha_{2}\right)$$and$$\left(\alpha_{3}, \alpha_{4}\right)$ as the ends of its one diameter is : (a) $$(x-2)(x-8)+(y-4)(y-16)=0$$ (b) $$(x-4)(x-8)+(y-2)(y-16)=0$$ (c) $$(x-2)(x-16)+(y-4)(y-8)=0$$ (d) $$(x-6)(x-8)+(y-5)(y-6)=0$$

Expert verified
(a) $$(x-2)(x-8)+(y-4)(y-16)=0$$
See the step by step solution

## Step 1: Compute the function values

Since $$f(1) = 2$$ and given $$f(x+y) = f(x) f(y)$$, we can obtain $$f(2) = f(1+1) = f(1)^2 = 2^2 = 4$$ and $$f(3) = f(2+1) = f(2) f(1) = 4 * 2 = 8$$. Similarly, find $$f(4) = f(3+1) = f(3) f(1) = 8 * 2 = 16$$. Hence, $$\alpha_{1} = f(1) = 2$$, $$\alpha_{2} = f(2) = 4$$, $$\alpha_{3} = f(3) = 8$$ and $$\alpha_{4} = f(4) = 16$$.

## Step 2: Apply the circle equation formula

Now, apply the formula $$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$$ for the circle with the given diameters $$(\alpha_{1}, \alpha_{2})$$ and $$(\alpha_{3}, \alpha_{4})$$ to obtain the circle equation. Which becomes $$(x-\alpha_1)(x-\alpha_3) + (y-\alpha_2)(y-\alpha_4) = 0$$. Substituting the computed function values we get $$(x-2)(x-8) + (y- 4)(y- 16) = 0$$.

## Step 3: Identify the correct option

Finally, compare your equation with the provided options to find the correct answer. According to our calculations, the correct option matches with option (a).

We value your feedback to improve our textbook solutions.

## Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features ## Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner 