Question

plzz solve q no 10 step by step ​

Answer 1

Length of tangent from (5, 3) to

x^2 +y^2 + 2x + ky +17 = 0 is 7

As length of tangent is

√ x^2 + y^2 + 2x + ky +17 where x,y denotes point from which tangent is drawn to point of contact

So put (5,3)

√25 + 9 + 10 + 3k +17 = 7

Squaring

34 +10 + 3k +17 = 49

61 + 3k -49 = 0

12 + 3k = 0

k = -4

Answer 2

Answer:-

a) -4

Step-by-step explanation:

Let (5,3) be (x,y)

Length in (x,y)=√(x²+y²+2gx+2py+c)

7=√(5²+3²+2×1×5+2×k×3+17)

7²=61+3k

3k=7²-61=-12

K=-4

HOPE IT HELPS U............