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Number of integral value(s) of k for which no tangent can be drawn from the point (k,k+2) to the circle x²+y² =4 is :

A. 0
B. 1
C. 2
D. 3

User Maxiss
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1 Answer

4 votes

Final answer:

There are 3 integral values of k (-2, -1, and 0) for which no tangent can be drawn from the point (k, k+2) to the circle x² + y² = 4, as the point must be inside or on the circle.

Step-by-step explanation:

To determine the number of integral values of k for which no tangent can be drawn from the point (k, k+2) to the circle + y² = 4, we need to consider the distance from the point to the center of the circle, which is at (0,0). A tangent can be drawn to a circle from a point outside it if the distance from the point to the center of the circle is greater than the radius of the circle.

The equation of the circle given is x² + y² = 4, which implies a radius of 2 units. For the point (k, k+2), we calculate the distance to the origin using the distance formula:

d = √((k-0)² + (k+2-0)²) = √(k² + k² + 4k + 4) = √(2k² + 4k + 4)

We now require that this distance d be less than or equal to 2 (the radius of the circle), as no tangent can be drawn if the point is inside or on the circle:

√(2k² + 4k + 4) ≤ 2

By squaring both sides of the inequality and simplifying:

2k² + 4k + 4 ≤ 4

2k² + 4k ≤ 0

Dividing everything by 2:

k² + 2k ≤ 0

This can be factored as:

k(k + 2) ≤ 0

The inequality is satisfied for values of k between -2 and 0, including -2 and 0. Thus, the integral values of k that satisfy this inequality are -2, -1, and 0. So, the answer is that there are 3 integral values of k for which no tangent can be drawn from the point to the circle.

User Ilan Lewin
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8.6k points