ExamBro
ExamBro
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:

  1. A \(185\)
  2. B \(85\)
  3. C \(95\)
  4. D \(195\)
Verified Solution

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 \(\left| {\frac{{ - 16 + 6 + 5}}{{\sqrt 5 }}} \right| = \sqrt {64 + 36 - C} \)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app