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What are the values of x y and k when the line y equals 3x plus 1 is a tangent to the circle x squared plus y squared equals k?
If y = 3x + 1 is a tangent to x² + y² = k (k > 0 since it is a square), then where they meet has a repeated root; they meet at:
x² + (3x + 1)² = k
→ x² + 9x² + 6x + 1 - k = 0
→ 10x² + 6x + (1 - k) = 0
This is the point of contact when it has a repeated root which is when the discriminant is zero, ie when:
6² + 4 × 10 × (1 - k) = 0
→ 36 + 40 - 40k = 0
→ 40k = 4
→ k = 1/10
I guess for x & y you mean the point where y = 3x + 1 is a tangent to x² + y² = k, ie the point of contact.
The value of k can now be substituted into the equation of the point of contact:
10x² + 6x + (1 - k) = 0
→ 10x² + 6x + (1 - 1/10) = 0
→ 10x² + 6x + 9/10 = 0
→ x² + 6x/10 + 9/100 = 0
→ (x + 3/10)²
→ The point of contact is when x = -3/10
→ y = 3× -3/10 + 1 = 9/10 + 1 = 1/10
→ point of contact is (-3/10, 1/10) with k = 1/10
What else can I help you with?
What are the possible values of k when the line y equals kx -2 is a tangent to the circle y equals x2 -8x plus 7?
If: y = kx -2 is a tangent to the curve (which is not a circle) of y = x^2 -8x +7 Then: kx -2 = x^2 -8x +7 Transposing and collecting like terms: (8x+kx) -x^2 -9 = 0 Using the discriminant: (8+k)^2 -4*-1*-9 = 0 Multiplying out the brackets and collecting like terms: 16k +k^2 +28 = 0 Factorizing the above: (k+2)(k+14) = 0 meaning k = -2 or k = -14 Therefore the possible values of k are -2 or -14
What are the values of x and y when x minus y equals 0 and x squared plus y squared equals 18?
(x, y) = (-3, -3) or (3, 3)
What are the possible values of k in the line y equals kx -2 which is tangent to the curve y equals x squared -8x plus 7?
The possible values for k are -2 and -14 because in order for the line to be tangent to the curve the discriminant must be equal to 0 as follows:- -2x-2 = x2-8x+7 => 6-x2-9 = 0 -14x-2 = x2-8x+7 => -6-x2-9 = 0 Discriminant: 62-4*-1*-9 = 0
What are the values of the variables when y equals x plus c is a tangent to the curve y equals 3 -x -5x squared?
-2
How do you prove in 3 different ways that the line of y equals x -4 is tangent to the curve of x squared plus y squared equals 8?
If: y = x-4 and y = x2+y2 = 8 then 2x2-8x+8 = 0 and the 3 ways of proof are: 1 Plot the given values on a graph and the line will touch the curve at one point 2 The discriminant of b2-4ac of 2x2-8x+8 must equal 0 3 Solving the equation gives x = 2 or x = 2 meaning the line is tangent to the curve
What are the possible values of k when the line y equals kx -2 is a tangent to the circle y equals x2 -8x plus 7?
If: y = kx -2 is a tangent to the curve (which is not a circle) of y = x^2 -8x +7 Then: kx -2 = x^2 -8x +7 Transposing and collecting like terms: (8x+kx) -x^2 -9 = 0 Using the discriminant: (8+k)^2 -4*-1*-9 = 0 Multiplying out the brackets and collecting like terms: 16k +k^2 +28 = 0 Factorizing the above: (k+2)(k+14) = 0 meaning k = -2 or k = -14 Therefore the possible values of k are -2 or -14
What are the values of k when the line of y equals kx -2 is a tangent to the curve of y equals x squared -8x plus 7?
If: y = kx -2 and y = x^2 -8x+7 Then the values of k work out as -2 and -14 Note that the line makes contact with the curve in a positive direction or a negative direction depending on what value is used for k.
What are the values of x and y when x minus y equals 0 and x squared plus y squared equals 18?
(x, y) = (-3, -3) or (3, 3)
What are the possible values of k in the line y equals kx -2 which is tangent to the curve y equals x squared -8x plus 7?
The possible values for k are -2 and -14 because in order for the line to be tangent to the curve the discriminant must be equal to 0 as follows:- -2x-2 = x2-8x+7 => 6-x2-9 = 0 -14x-2 = x2-8x+7 => -6-x2-9 = 0 Discriminant: 62-4*-1*-9 = 0
What are the values of the variables when y equals x plus c is a tangent to the curve y equals 3 -x -5x squared?
-2
How do you prove in 3 different ways that the line of y equals x -4 is tangent to the curve of x squared plus y squared equals 8?
If: y = x-4 and y = x2+y2 = 8 then 2x2-8x+8 = 0 and the 3 ways of proof are: 1 Plot the given values on a graph and the line will touch the curve at one point 2 The discriminant of b2-4ac of 2x2-8x+8 must equal 0 3 Solving the equation gives x = 2 or x = 2 meaning the line is tangent to the curve
Why can tangent values be greater than 1 but sine and cosine values cannot be greater than 1?
Sine and cosine cannot be greater than 1 because they are the Y and X values of a point on the unit circle. Tangent, on the other hand, is sine over cosine, so its domain is (-infinity,+infinity), with an asymptote occurring every odd pi/2.
What is sin squared x equals cos squared minus 2 sin x?
Since the word 'equals' appears in your questions it might be what is called a trigonometric identity, in other words a statement about a relationship between various trigonometric values.
What are the possible values of k when the equation x squared plus 2kx plus 81 equals 0 has equal roots?
Using the discriminant the possible values of k are -9 or 9
Where do the curves of y equals 4x squared -2x -1 and y equals -2x squared plus 3x plus 5 intersect?
They intersect at points (-2/3, 19/9) and (3/2, 5) Solved by combining the two equations together to equal nought and then using the quadratic equation formula to find the values of x and substituting these values into the equations to find the values of y.
What do the asymptotes represent when you graph the tangent function?
When you graph a tangent function, the asymptotes represent x values 90 and 270.
What is the point of contact when the line y -3x -5 equals 0 meets the circle x squared plus y squared -2x plus 4y -5 equals 0 and how would you prove that it is a tangent line showing work?
The line y = -3x - 5 meets the circle x² + y² -2x + 4y - 5 = 0 at the points where the x and y values solve both equations:y -3x - 5 = 0x² + y² - 2x + 4y - 5 = 0Rearrange (1) to make y the subject and substitute for y in (2) and solve for x, then use the value(s) found to solve for y (using 1):y - 3x - 5 = 0→ y = 3x + 5→ x² + (3x + 5)² - 2x + 4(3x + 5) - 5 = 0→ x² + 9x² + 30x + 25 - 2x + 12x + 20 - 5 = 0→ 10x² + 40x + 40 = 0→ x² +4x + 4 = 0→ (x +2)² = 0→ x = -2This is a repeated root of the quadratic;the quadratic only has one solution;thus the line and the circle meet at only one point;therefore the line is a tangent to the circle.→ y - 3×-2 - 5 = 0→ y + 6 - 5 =→ y = -1→ the line and the circle meet at the point (-2, -1).The line and the circle meet at the point (-2, -1) and the line is a tangent to the circle.
