Find the point in the first quadrant where the line y = 4 x intersects a circle of radius 6 centered at the origin.?
How do I go about in solving this problem? So since, its centered at the origin I would use: x^2+y^2=r^2 I'd get: x^2+(4x)^2=36 17x^2=36 Divide both sides by 17 x^2=36/17 Square both sides x=+-sqrt(36/17) Idk why I'm getting it wrong when I plug this in. Can someone please explain what I'm doing wrong? If I did this incorrectly, can someone please show me the steps? Please and thank you!
At what point in the second quadrant does the line y=2x+5 intersect a circle with radius 3 and center (-2,0)?
A circle is written in the form (x - h)² + (y - k)² = r², where r = radius and (h,k) = coordinates of the center. So your circle is: (x + 2)² + y² = 9 And you want to know where that intersects the line y = 2x + 5, so just substitute the value of y in the equation of the line in the circle's equation and solve for x. (x + 2)² + (2x + 5)² = 9 x² + 4x + 4 + 4x² + 20x + 25 = 9 5x² + 24x + 29 = 9 5x² + 24x + 20 = 0 Now to solve that quadratic you have to use the quadratic formula or complete the square. Utilizing the later: x² + 24x / 5 + 4 = 0 (I divided both sides by 5) (x + 24/10)² = -4 + (24/10)² (x + 12/5)² = -4 + 144/25 (x + 12/5)² = 44 / 25 x + 12/5 = ± 2√(11) / 5 x = [-12 ± 2√(11)] / 5 x ≈ -3.73, -1.073 y = 2x + 5 = 2(-3.73) + 5 => a negative number so not in 2nd quadrant y = 2x + 5 = 2(-1.073) + 5 => 2.854 So the point of intersection is ≈ (-1.073, 2.854).
At what point in the first quadrant does the line with equation y=x+2 intersect the circle with radius r=6 and the center (-1,0)?
(x, y) in Q1 means that x and y are both positive numbers. Circle: (x+1)² + y² = 36 Line: y = x+2 Substitute for y in the circle: (x+1)² + (x+2)² = 36 x² + 2x + 1 + x² + 4x + 4 = 36 2x² + 6x - 31 = 0 x = [-6 ± √284]/4 Discard the negative solution since x > 0 x = (-3 + √71)/2 y = x+2 = (1+7√2)/4 (x, y) = [(-3+√71)/2, (1+√71)/2] ≈ (2.713, 4.713)
Trig question about points in first quadrant?
How do i find an individual point in a given quadrant? for example at what point in the first quadrant does the line with equation y=2x+5 intersect a circle with radius 3 and center (0,5)? I know that the equation of the circle given this information is (x-0)^2+(y-5)^2=3^2 but where do i go from here? How do i find the point given the line equation? help pls!
How can you find the equation to the circle whose centre is in the first quadrant and radius 4, given that it touches the x axis and the line 4x-3y?
As per your question, the circle touches both y=(3/4)x and y=0, Slope of the line y=3/4 x is tan^(-1)(3/4)=75 degreesThat means the angle between the tangents from origin to the circle is 75 degreesLet tan(theta)=3/4=>2tan(theta/2)/1-tan^2(theta/2) = 3/4=>tan(theta/2)=1/3 We also know that tan(theta/2)=r/x, x is the x coordinate of the center of circle=>4/x=1/3=>x=12It is obvious that the y coordinate will be the radius of the circle Hence the equation of circle will be(X-12)^2 + (y-4)^2 = 16
A circle C has center at the origin and radius 8. Another circle K has a diameter with one end at the origin...?
A circle C has center at the origin and radius 8. Another circle K has a diameter with one end at the origin and the other end at the point (0,19).(0,19). The circles C and K intersect in two points. Let P be the point of intersection of C and K which lies in the first quadrant. Let (r,θ) be the polar coordinates of P, chosen so that r is positive and 0≤θ≤2. Find r and θ.
Circle and Line Intersecting Problem.. NEED HELP!!!!?
Recall the equation for a circle with center (h,k) and radius r. At what point in the first quadrant does the line with equation y = 1.5 x + 3 intersect the circle with radius 4 and center (0, 3)? Enter your answer correct to 3 decimal places . What I did was put the circle information into its equation making it: Circle Equation: (x-0)^2+(y-3)^2=4^2 Line Equation: y=1.5x+3 Now I'm stuck with two equations and don't know what to do next. Need help ASAP!
Find the radius of largest circle touching coordinate axes and 3x+4y=12?
I take it by “touch” you mean “tangent to”. Otherwise, the circle can be as big as you like, tangent to 3x+4y=12 and with center at (1000,1000) for example. It will cross the x and y axes twice each.So tangent it is. Because the third line crosses both axes at points with a positive coordinate, everything happens in the first quadrant. Because the circle must touch both axes, its center must be on the line in the first quadrant that bisects their angle, that is, along the line y=x.The bisector of the angle between the third line and the x axis must also pass through the center. So the question boils down to figuring out the equation of that bisector. The line segment from (4,0) to (0,3) along that third line has length 5. Take a vector of length 5 out along the x axis from (4,0) to (9,0). Add the two vectors [5,0] and [-4,3] (which have equal length) to get [1,3]. The equation of the line through (4,0) and parallel to that vector is y=3(x-4). Solve the system of equations y=x, y=3(x-4) by setting y=x from the first equation into the second: x=3(x-4). That works out to x=6, and of course then y=6. The center of the circle is (6,6) and because the circle must touch the x and y axis at a tangent, it has radius 6. The answer is 6, and the hardest algebra was multiplying out 3(x-4)=3x-12 and then x=3x-12 implies 0=2x-12 implies 2x=12 and the hardest arithmetic was multiplying 3*3+4*4=25=5*5, that or dividing 12 by 2.Sometimes, a little geometry takes the place of a fair bit of algebra.