0
What is the tangent line equation that touches the circle of x squared plus y squared -6x plus 4y -7 equals 0 at the point of 1 2 on the Cartesian plane showing work?
Wiki User
∙ 7y ago
The line tangent to a circle is perpendicular to the radius from the centre of the circle to the point of contact.
A circle with centre (X, Y) and radius r has equation:
(x - X)² + (y - Y)² = r²
Re-arranging x² + y² -6x + 4y -7 = 0 into this form allows us to find the centre of the circle. This requires completing the square in both x and y:
x² + y² - 6x + 4y - 7 = 0
→ x² -6x + 9 - 9 + y² + 4y + 4 - 4 - 7 = 0
→ (x - 3)² - 9 + (y + 2)² - 4 - 7 = 0
→ (x - 3)² + (y + 2)² - 20 = 0
→ (x - 3)² + (y + 2)² = 20
→ the centre of the circle is at (3, -2).
The slope of the radius between this centre and the point of touch at (1, 2) is given by:
m = change_in_y / change_in_x = (2 - -2)/(1 - 3) = 4/-2 = -2
The slope m' of a line perpendicular to a line of slope m is such that mm' = -1
→ m' = -1/m
→ the slope of the tangent m' is m' = -1/-2 = 1/2
The equation of a line through point (x0, y0) with slope m is given by:
y - y0 = m(x - x0)
Thus the equation of the tangent (with slope 1/2) to the circle at (1, 2) is:
y - 2 = ½(x - 1)
→ 2y - 4 = x - 1
→ 2y = x + 3
→ 2y - x = 3
(or y = x/2 + 1.5)
Wiki User
∙ 7y ago
Add your answer:
Earn +20 pts
What is the equation of a circle whose diameter endpoints are at 2 -3 and 8 7 on the Cartesian plane and what are the tangent equations touching its endpoints showing work?
Endpoints: (2, -3) and (8, 7)Centre of circle: (5, 2)Radius of circle is the square root of 34Equation of the circle: (x-5)^2 +(y-2)^2 = 34Slope of radius: 5/3Slope of tangents: -3/51st tangent equation: y--3 = -3/5(-2) => 5y = -3x-92nd tangent equation: y-7 = -3/5(x-8) => 5y = -3x+59
The line y 4xk is a tangent of the circle x squared y squared 17 Condition for this equation to have coincident roots?
A line can have at most one root and so coincident roots for a line are not possible.
How do you find the equation of a parabola if you know the equation of the tangent that touches it?
You need more than one tangent to find the equation of a parabola.
What is the point of contact when the tangent line 4x plus 9y plus 5 equals 0 touches the circle 2x squared plus 2y squared -8x -5y -1 equals 0 on the Cartesian plane showing work?
Equations: 4x+9y+5 = 0 and 2x^2 +2y^2 -8x -5y -1 = 0 Divide all terms in the 1st equation by 4 and in the 2nd equation by 2 So: x+2.25y+1.25 = 0 and x^2 +y^2 -4x -2.5y -0.5 = 0 If: x+2.25y+1.25 = 0 Then: x = -2.25y-1.25 If: x^2 +y^2 -4x -2.5y -0.5 = 0 Then: (-2.25y -1.25)^2 +y^2 -4(-2.25y -1.25) -2.5y -0.5 = 0 Removing brackets and collecting like terms: 6.0625y^2 +12.125y +6.0625 = 0 Using the quadratic equation formula gives y equal values of -1 By substitution the tangent makes contact with the circle at (1, -1) on the Cartesian plane
What is the length of the tangent line from the point 0 0 to a point where it touches the circle x2 plus y2 plus 4x -6y plus 10 equals 0 on the Cartesian plane?
Equation of the circle: x^2 +y^2 +4x -6y +10 = 0 Completing the squares: (x+2)^2 +(y-3)^2 = 3 Radius of the circle: square root of 3 Center of circle: (-2, 3) Distance from (0, 0) to (-2, 5) = sq rt of 13 which is the hypotenuse of right triangle. Using Pythagoras' theorem : distance squared - radius squared = 10 Therefore length of tangent line is the square root of 10 Note that the tangent of a circle meets its radius at right angles.
What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?
Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius
What is the tangent equation of the circle 2x squared plus 2y squared -8x -5y -1 that touches the point of 1 -1 on the Cartesian plane?
If you mean: 2x^2 +2y^2 -8x -5y -1 = 0 making contact at (1, -1) Then the tangent equation in its general form works out as: 4x+9y+5 = 0
What is the tangent equation that touches the circle of x squared plus y squared -8x -y plus 5 equals 0 at the point of 1 2 on the Cartesian plane showing work?
Equation of circle: x^2 +y^2 -8x -y +5 = 0Completing the squares: (x-4)^2 +(y-0.5)^2 = 11.25Centre of circle: (4, 0.5)Slope of radius: -1/2Slope of tangent: 2Equation of tangent: y-2 = 2(x-1) => y = 2xNote that the above proves the tangent of a circle is always at right angles to its radius
What is the equation of the tangent line that touches the circle x squared plus 10x plus y squared -2y -39 equals 0 at the point of 3 2 on the Cartesian plane showing work?
First find the slope of the circle's radius as follows:- Equation of circle: x^2 +10x +y^2 -2y -39 = 0 Completing the squares: (x+5)^2 + (y-1)^2 -25 -1 -39 = 0 So: (x+5)^2 +(y-1)^2 = 65 Centre of circle: (-5, 1) and point of contact (3, 2) Slope of radius: (1-2)/(-5-3) = 1/8 which is perpendicular to the tangent line Slope of tangent line: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26 Tangent equation in its general form: 8x+y-26 = 0
What is the equation of the tangent line that touches the circle x squared plus y squared -6x plus 4y equals 0 at the point of 6 -4 on the Cartesian plane?
Equation: x² + y² -6x +4y = 0 Completing the squares: (x-3)² + (y+2)² = 13 Centre of circle: (3, -2) Contact point: (6, -4) Slope of radius: -2/3 Slope of tangent: 3/2 Tangent equation: y - -4 = 3/2(x-6) => 2y - -8 = 3x-18 => 2y = 3x-26 Tangent line equation in its general form: 3x-2y-26 = 0
What is the tangent equation that touches the circle 2x squared plus 2y squared -8x -5y -1 equals 0 at the point of 1 -1 on the Cartesian plane showing work?
Circle equation: 2x^2 +2y^2 -8x -5y -1 = 0 Divide all terms by 2: x^2 +y^2 -4x -2.5y -0.5 = 0 Complete the squares: (x-2)^2 +(y-1.25)^2 -4 -1.5625 -1 = 0 So: (x-2)^2 +(y-1.25)^2 = 6.0625 Centre of circle: (2, 1.25) Contact point: (1, -1) Slope of radius: (-1-1.25)/(1-2) = 9/4 Slope of tangent line: -4/9 Tangent equation: y- -1 = -4/9(x-1) => 9y--9 = -4x+4 => 9y = -4x-5 Tangent equation in its general form: 4x+9y+5 = 0
What is the tangent line equation that touches the curve of 2x squared plus 2y squared -8x -5y -1 equals 0 at the point of 1 and -1 on the Cartesian plane?
Equation of the curve: 2x^2 +2y^2 -8x -5y -1 = 0 Divide all terms by two: x^2 +y^2 - 4x -2.5y -0.5 = 0 Completing the squares: (x-2)^2 +(y-1.25)^2 = 6.0625 Centre of circle: (2, 1.25) Contact point: (1, -1) Slope of radius: 9/4 Slope of tangent: -4/9 Tangent equation: y--1 = -4/9(x-1) => 9y --9 = -4x+4 => 9y = -4x-5 Tangent line equation in its general form: 4x+9y+5 = 0
What is the equation of the tangent line that touches the circle of 2x squared plus 2y squared -8x -5y -1 equals 0 at the point of 1 -1 on the Cartesian plane?
Equation of circle: 2x^2 +2y^2 -8x -5y -1 = 0 Divide all terms by two: x^2 +y^2 -4x -2.5 -0.5 = 0 Completing the squares: (x-2)^2 + (y-1.25)^2 = 97/16 Point of contact: (1, -1) Centre of circle: (2, 1.25) Slope of radius: 9/4 Slope of tangent: -4/9 Equation of tangent: y--1 = -4/9(x-1) => 9y--9 = -4x+4 => 9y = -4x-5 Tangent equation in its general form: 4x+9y+5 = 0
What is the equation of a circle whose diameter endpoints are at 2 -3 and 8 7 on the Cartesian plane and what are the tangent equations touching its endpoints showing work?
Endpoints: (2, -3) and (8, 7)Centre of circle: (5, 2)Radius of circle is the square root of 34Equation of the circle: (x-5)^2 +(y-2)^2 = 34Slope of radius: 5/3Slope of tangents: -3/51st tangent equation: y--3 = -3/5(-2) => 5y = -3x-92nd tangent equation: y-7 = -3/5(x-8) => 5y = -3x+59
How do you prove that the line y equals x-4 is tangent to the curve of x squared plus y squared equals 8?
equation 1: y = x-4 => y2 = x2-8x+16 when both sides are squared equation 2: x2+y2 = 8 Substitute equation 1 into equation 2: x2+x2-8x+16 = 8 => 2x2-8x+8 = 0 If the discriminant of the above quadratic equation is zero then this is proof that the line is tangent to the curve: The discriminant: b2-4ac = (-8)2-4*2*8 = 0 Therefore the discriminant is equal to zero thus proving that the line is tangent to the curve.
The line y 4xk is a tangent of the circle x squared y squared 17 Condition for this equation to have coincident roots?
A line can have at most one root and so coincident roots for a line are not possible.
What is the equation of the tangent line that touches the circle x squared plus y squared -8x -16y -209 equals 0 at a coordinate of 21 and 8?
Circle equation: x^2 +y^2 -8x -16y -209 = 0 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Centre of circle: (4, 8) Radius of circle 17 Slope of radius: 0 Perpendicular tangent slope: 0 Tangent point of contact: (21, 8) Tangent equation: x = 21 passing through (21, 0)
