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How do you find the angle of a slope?
its the tangent of the angle the slope makes with the x-axis
What does tangent to the y axis mean?
Normally a straight line is a tangent to a curved line but, presumably, that relationship can be reversed. So a tangent to the y axis would be a curve that just touches the y axis but does not cross it - at least, not at the point of tangency.
What are the tangent equations to the circle x2 plus y2 -6x plus 4y plus 5 equals 0 at the points where they meet the x axis?
Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares (x -3)^2 +(y +2)^2 = 8 Centre of circle: (3, -2) Radius of circle: square root of 8 Points of contact are at: (1, 0) and (5, 0) where the radii touches the x axis Slope of 1st tangent line: 1 Slope of 2nd tangent line: -1 Equation of 1st tangent: y -0 = 1(x -1) => y = x -1 Equation of 2nd tangent: y -0 = -1(x -5) => y = -x +5
Find the differential equation of all circles tangent to y-axis?
Let the circle with centre (a, b) be tangent to the y-axis. Then, the radius of the circle must be b. Therefore the equation of the circle is (x - a)2 + (y - b)2 = b2 or x2 - 2ax + a2 + y2 - 2by = 0 Then 2x - 2a + 2ydy/dx - 2bdy/dx = 0 ie x - a + ydy/dx - bdy/dx (y - b)dy/dx = a - x so dy/dx = (a - x)/(y - b) or -(x - a)/(y - b)
What is the distance from a point on the x axis to the center of the circle x2 plus y2 -2x -6y plus 5 equals 0 when the tangent of the circle is at 3 4?
If I understand your question you are asking for the distance from the point of intersection of the x-axis and the tangent to the circle at (3, 4) to the centre of the circle.To solve this you need to:Find the centre of the circle (X, Y) - a circle with equation (x - X)² + (y - Y)² = r² has a centre of (X, Y) and radius r;Find the slope m of the radius from the centre of the circle and the point of contact of the tangent;Use the slope m of the radius to calculate the slope m' of the tangent - the slopes of two perpendicular lines is such that mm' = -1;Use the slope-point equation of a line to find the equation of the tangent - a line with slope m through point (X, Y) has equation: y - Y = m(x - X);Find the point where this line crosses the x-axisUse Pythagoras to find the distance from this point to the centre of the circle.Have a go before reading the solution belowHint: You need to complete the squares in x and y to rearrange the equation for the circle into the form in step 1.---------------------------------------------------------------------1. Find the centre of the circle x² + y² - 2x - 6y + 5 = 0x² + y² - 2x - 6y + 5 = 0→ x² - 2x + y² - 6y + 5 = 0→ (x - 2/2)² - (2/2)² + (y - 6/2)² - (6/2)² + 5 = 0→ (x - 1)² - 1 + (y - 3)² - 9 + 5 = 0→ (x - 1)² + (y - 3)² = 5→ centre of circle is at (1, 3)2. Find the slope of the radius to (3, 4)slope = change_in_y/change_in_x→ m = (4 - 3)/ (3 - 1)= 1/23. Calculate the slope of the tangentmm' = -1→ m' = -1/m= -1(1/2)= -2→ slope of tangent is -24. Find the equation of the tangenty - 4 = -2(x - 3)→ y - 4 = -2x + 6→ y + 2x = 105. Find the point where the tangent crosses the x-axisx-axis is the line y = 0→ y + 2x = 10→ 0 + 2x = 10→ 2x = 10→ x = 5→ tangent crosses x-axis at (5, 0)6. Find the distance using Pythagorasdistance = √(difference_in_x² + difference_in_y²)= √((5 - 1)² + (3 - 0)²)= √(4² + 3²)= √(16 + 9)= √25= 5→ The distance from the intersection of the x-axis and the tangent to the circle x² + y² - 2x - 6y + 5 = 0 at (3, 4) to the centre of that circle is 5 units.
