JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If the in centre of an equilateral triangle is \((1, 1)\) and the equation of its one side is \(3x + 4y + 3\,= 0\), then the equation of the circumcircle of this triangle is
- A \({x^2} + {y^2} - 2x - 2y - 14 = 0\)
- B \({x^2} + {y^2} - 2x - 2y - 2 = 0\)
- C \({x^2} + {y^2} - 2x - 2y +2 = 0\)
- D \({x^2} + {y^2} - 2x - 2y - 7 = 0\)
Answer & Solution
Correct Answer
(A) \({x^2} + {y^2} - 2x - 2y - 14 = 0\)
Step-by-step Solution
Detailed explanation
Let radius of circumircles be According to the question, \(\frac{r}{2} = \frac{{10}}{5} \Rightarrow r = 4\) So equation of Required circle is \({\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 16\) \( \Rightarrow {x^2} + {y^2} - 2x - 2y - 14 = 0\)
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