The number of integral values of K for which x^2+ y ^2+ k x+(1- k ) y +5=0 represents a circle whose radius cannot exceeds 5, are

Step-by-step solution

The circle equation is x^2 + y^2 + Kx + (1-K)y + 5 = 0 . Comparing with the standard form x^2 + y^2 + 2gx + 2fy + c = 0 , we identify 2g = K , 2f = 1-K , and c = 5 . The radius squared is r^2 = g^2 + f^2 - c = ( K 2 )^2 + ( 1-K 2 )^2 - 5 = 2K^2 - 2K + 1 4 - 5 . Given r 5 , it follows that r^2 25 . 2K^2 - 2K - 119 0 The roots of 2K^2 - 2K - 119 = 0 are K = 2 956 4 2 30.919 4 , yielding K₁ -7.22975 and K₂ 8.22975 . The inequality holds for K between these roots: -7.22975 K 8.22975 . The integral values within this ra