JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If the point \((1, 4)\) lies inside the circle \(x^2 + y^2-6x - 10y + p = 0\) and the circle does not touch or intersect the coordinate axes, then the set of all possible values of \(p\) is the interval
- A \((0,25)\)
- B \((25, 39)\)
- C \((9, 25)\)
- D \((25, 29)\)
Answer & Solution
Correct Answer
(D) \((25, 29)\)
Step-by-step Solution
Detailed explanation
The equation of circle is \({x^2} + {y^2} - 6x - 10y + P = 0\,\,\,\,\,......\left( i \right)\) \({\left( {x - 3} \right)^2} + {\left( {y - 5} \right)^2} = {\left( {\sqrt {34 - P} } \right)^2}\) center \((3,5)\) and radius \('r' = {\left( {\sqrt {34 - P} } \right)^2}\) If cicrle…
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