The equation of a circle is given by \(x^{2} + y^{2} - 4x - 2y - 3\). Find the radius and the coordinates of its centre.
Question 1 Report
The equation of a circle is given by \(x^{2} + y^{2} - 4x - 2y - 3\). Find the radius and the coordinates of its centre.
Answer Details
The equation of a circle in standard form is \((x-a)^2 + (y-b)^2 = r^2\), where \((a,b)\) are the coordinates of the center and \(r\) is the radius. We need to rewrite the given equation in this standard form by completing the square for both the \(x\) and \(y\) terms.
\begin{align*}
x^{2} + y^{2} - 4x - 2y - 3 &= 0\\
(x^{2} - 4x) + (y^{2} - 2y) &= 3\\
(x^{2} - 4x + 4) + (y^{2} - 2y + 1) &= 3 + 4 + 1\\
(x-2)^{2} + (y-1)^{2} &= 8
\end{align*}
Comparing with the standard form, we see that the center of the circle is \((2,1)\) and the radius is \(\sqrt{8} = 2\sqrt{2}\).
Therefore, the answer is: the radius of the circle is \(2\sqrt{2}\) and the coordinates of its center are \((2,1)\).
Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning