JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If the tangent at \(\left( {1,7} \right)\) to the curve \({x^2} = y - 6\) touches the circle \({x^2} + {y^2} + 16x + 12y + c = 0\) then the value of \(c\) is:
- A \(185\)
- B \(85\)
- C \(95\)
- D \(195\)
Answer & Solution
Correct Answer
(C) \(95\)
Step-by-step Solution
Detailed explanation
Equation of tangert at \((1,7)\) to \({x^2} = y - 6\) is \(2x - y + 5 = 0\) Now, perpentdicular from center \(O(-8,-6)\)to \(2x - y + 5 = 0\) should be equal to radium of the circle \(\left| {\frac{{ - 16 + 6 + 5}}{{\sqrt 5 }}} \right| = \sqrt {64 + 36 - C} \)…
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