Let c and d be vectors satisfying c ( i - 2 j + 2 k ) = ( i - 2 j + 2 k ) d . Let u = c + d such that u i > 0 . If the equation x^2 + y^2 + ( u i )x + ( u j )y + c = 0 represents a circle which touches the straight line 3x + 4y - 30 = 0 at the point (2, 6) , then the value of | c + d |^2 + c is:

Step-by-step solution

Given c ( i - 2 j + 2 k ) = ( i - 2 j + 2 k ) d . This can be written as ( c + d ) ( i - 2 j + 2 k ) = 0 . Thus, u = c + d = ( i - 2 j + 2 k ) for some scalar . Since u i > 0 , we have > 0 . The equation of the circle is x^2 + y^2 + x - 2 y + c = 0 . The center of this circle is (- 2 , ) . The circle touches the line 3x + 4y - 30 = 0 at (2, 6) . The normal to the circle at the point of tangency must pass through the center. The slope of the tangent line is - 3 4 , so the slope of the normal is 4 3 . Equation of the