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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If two vertices of an equilateral triangle are \(A (- a, 0)\) and \(B ( a, 0),\,a > 0,\) and the third vertex \(C\) lies above \(x-\) axis then the equation of the circumcircle of \(\Delta ABC\) is
- A \(3{x^2} + 3{y^2} - 2\sqrt 3 ay = 3{a^2}\)
- B \(3{x^2} + 3{y^2} - 2ay = 3{a^2}\)
- C \({x^2} + {y^2} - 2ay = {a^2}\)
- D \({x^2} + {y^2} - \sqrt 3ay = {a^2}\)
Answer & Solution
Correct Answer
(A) \(3{x^2} + 3{y^2} - 2\sqrt 3 ay = 3{a^2}\)
Step-by-step Solution
Detailed explanation
Let \(C=(x,y)\) Now, \( \Rightarrow {\left( {x + a} \right)^2} + {y^2} = {\left( {x + a} \right)^2} + {y^2} = {\left( {2a} \right)^2}\) \( \Rightarrow {x^2} + 2ax + {a^2} + {y^2} = 4{a^2}\,\,\,\,\,\,\,.......\left( i \right)\) and…
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