Find values of k so that hte line 2x+ky=6 will be tangent to the circle (x-1)^2 + (y+6)^2 =20.?

Answer 1

From the line, we have:
x-1 = 2 - ky/2

Plug this into the circle's equation to get y values of the intersection:
(2 - ky/2)^2 + (y+6)^2 = 20
4 + k^2 * y^2 / 4 - 2ky + y^2 + 36 + 12y = 20
(1+k^2/4) y^2 + (12-2k)y + 20 = 0

Now for a tangent, we must have only one y value for the intersection. In other words, the discriminant of quadratic formula must be 0:
(12-2k)^2 - 4(1+k^2/4)20 = 0
(6-k)^2 = 5(4+k^2)
k^2 - 12k + 36 = 20 + 5k^2
4k^2 + 12k - 16 = 0
k^2 + 3k - 4 = 0
(k+4)(k-1) = 0

So k = -4 or k = 1

k = -4 => intersection point is (-1, -2)
and
k = 1 => intersection point is (5, -4)

Answer 2

Given the circle (x - 1)² + (y + 6)² = 20 and the line 2x + ky = 6, find the values of k so that the line will be tangent to the circle.

The center of the circle is the point (1, -6). The radius of the circle is √20. The distance from the center of the circle to the line must also be √20.

The equation of the line is:

2x + ky = 6
2x + ky - 6 = 0

The distance from the center of the circle (1, -6) to the line is:

D = | 2*1 - 6k - 6 | / √(2² + k²) = √20

| - 6k - 4 | = (√20)*√(2² + k²)

Square both sides.

(6k + 4)² = 20(4 + k²)
36k² + 48k + 16 = 80 + 20k²
16k² + 48k - 64 = 0
k² + 3k - 4 = 0
(k + 4)(k - 1) = 0

k = -4, 1

Answer 3

x² + y² + D'x + E'y + F' = 0 (a million)² + (5)² + D'(a million) + E'(5) + F' = 0 a million. D' + 5E' + F' = -26 x² + y² + D'x + E'y + F' = 0 (-2)² + (3)² + D'(-2) + E'(3) + F' = 0 2. -2d' + 3E' + F' = -13 x² + y² + D'x + E'y + F' = 0 (2)² + (-a million)² + D'(2) + E'(-a million) + F' = 0 3. 2d' - E' + F' = -5 the perfect thank you to sparkling up those "concurrently" is to become conscious of a few linear combos that artwork--and it is something that takes extra special genius to end. So instead i'll merely make a matrix: a million. D' + 5E' + F' = -26 2. -2d' + 3E' + F' = -13 3. 2d' - E' + F' = -5 |...a million...5...a million...|...-26...| |..-2...3...a million...|...-13...| |...2..-a million...a million...|....-5....| i'm already annoying with regards to the 0.33 column... upload row 2 to row 3 |...a million...5...a million...|...-26...| |..-2...3...a million...|...-13...| |...0...2...2...|...-18...| upload 2 circumstances row a million to row 2 |...a million...5...a million...|...-26...| |...0..13..3...|...-sixty 5...| |...0...2...2...|...-18...| we could divide row 3 via by ability of two |...a million...5...a million...|...-26...| |...0..13..3...|...-sixty 5...| |...0...a million...a million...|....-9....| Then swap rows 2 and 3 |...a million...5...a million...|...-26...| |...0...a million...a million...|....-9....| |...0..13..3...|...-sixty 5...| Multiply -5(row 2) + row1?row a million and -13(row 2) + row 3?row 3 |...a million...0..-4...|...19....| |...0...a million...a million...|....-9....| |...0...0 .-10.|... fifty two...| Divide row 3 by ability of -10 |...a million...0..-4...|...19....| |...0...a million...a million...|....-9....| |...0...0...a million...|...-26/5| Now we choose 4(row 3) + row a million ? row a million and -(row 3) + row 2?row 2 |...a million...0...0...|...-9/5..| |...0...a million...0...|.-19/5..| |...0...0...a million...|.-26/5..| Phew, ok it relatively is in all possibility the 1st time you have ever seen a matrix row decreased in our existence. a thank you to interpret it is: our first a million corresponds to our first variable, our 2d a million to our 2d, and our 0.33 a million to our 0.33. So: D' = -9/5, E' = -19/5, F' = -26/5. i'm hoping that helped, there are different the right thank you to try this, it is (have faith it or no longer) merely bar none the quickest for basic numbers. I infant stepped it yet row decreasing a matrix is readily for a three x 3 equipment of equations.

Answer 4

for 2x+ky=6 [1] to be tangent to (x-1)^2 + (y+6)^2 =20 [2] we need the line to cross the circle in one point and one point only.

write:
2x+ky=6 as x=(6-ky)/2
now substitute for x in [2] and expand:
you will find a polynomial of the form:
a(k)y^2+b(k)y+c(k) = 0
where a(k), b(k) and c(k) are coefficient which depend on k.

We only allow 1 solution for 1, which implies that Delta = b^2-4*a*c = 0

or in our case:
b(k)^2-4*a(k)c(k) = 0
This is again a 2nd-order polynomial of the form:
pk^2+qk+r=0, with p,q,r constants this time.
The solutions for k are then given by:

k1 = (-q+(q^2-4*p*r)^0.5)/(2p)
and
k2 = (-q-(q^2-4*p*r)^0.5)/(2p)

and there you go, you have your solutions for k. Sorry I didn t do the calculations. You only need to plug in the numbers now.

Answer 5

The distance of the center to the line must be = radius
C(1,-6)
Distance= 2-6k-6=+-sqrt(4+k^2)*sqrt20
so
(-6k-4)^2= 20*(4+k^2) and 36k^2+48k+16=80+20k^2
16k^2+48k-64=0
k^2+3k-4=0 so k=((-3+-sqrt(9+16))/2
k=1 and k= -4

Answer 6

TAN,
x^2=y^2=r^2
[x-1]^2+[y=6]^2=20[r=root of 20]
x=1
y=-6
2x+ky=6
2[1]-6k=6
2-6=6k
-4=6k
k=-2/3