Question

For which value(s) of the constant k is the circle x² + (y − k)² = 16 tangent to the line y = 3?

Answer 1

Answer:

-1, 7

Step-by-step explanation:

Equation of the circle:

x² + (y − k)² = 16

When the circle intersects y = 3:

x² + (3 − k)² = 16

x² + 9 − 6k + k² = 16

x² = 7 + 6k − k²

x = ±√(7 + 6k − k²)

For the circle to be tangent to the line, it can only intersect at one point.  If x has only one value, then:

√(7 + 6k − k²) = -√(7 + 6k − k²)

2√(7 + 6k − k²) = 0

7 + 6k − k² = 0

k² − 6k − 7 = 0

(k − 7) (k + 1) = 0

k = -1, 7

The two values of k are -1 and 7.

Answer 2

Answer:

Step-by-step explanation:Let us find points of intersection of line  

3

x

+

4

y

−

k

=

0

and circle  

x

2

+

y

2

=

16

. We can do this by putting value of  

y

from first equation i.e.  

y

=

k

−

3

x

4

and we get

x

2

+

(

k

−

3

x

)

2

16

=

16

or  

16

x

2

+

k

2

+

9

x

2

−

6

k

x

=

256

i.e.  

25

x

2

−

6

k

x

+

k

2

−

256

=

0

This would give two values of  

x

and corresponding two values of  

y

i.e. two points. But tangent cuts the circle in only at one point. This will be so when discriminant is zero i.e.

(

−

6

k

)

2

−

4

⋅

25

⋅

(

k

2

−

256

)

=

0

or  

−

64

k

2

+

25600

=

0

or  

k

=

±

20

graph{(x^2+y^2-16)(3x+4y-20)(3x+4y+20)=0 [-10, 10, -5, 5]}