Question
Easy
Centre and radius of the circle $2x^{2}+2y^{2}+16x+20y-16=0$ be
1
(-4, -5); 7
2
(-8,-10); 14
3
(4, 5); 7
4
(8, 10); 14
Question Details
Time to Solve: 12
Exam: HTET
Level/Paper: Level 3
Chapter:
Topic:
Correct Answer
Option A
Explanation
To determine the center and radius of the circle given by the equation \(2x^{2} + 2y^{2} + 16x + 20y - 16 = 0\), we first need to rewrite the equation in the standard form of a circle, \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. ### Step-by-step Explanation: 1. Simplify the Equation: Divide the entire equation by…Read More
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