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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?
Wiki User
∙ 6y ago
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-axis
- Use Pythagoras to find the distance from this point to the centre of the circle.
Have a go before reading the solution below
Hint: You need to complete the squares in x and y to rearrange the equation for the circle into the form in step 1.
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1. Find the centre of the circle x² + y² - 2x - 6y + 5 = 0
x² + 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)
= 1/2
3. Calculate the slope of the tangent
mm' = -1
→ m' = -1/m
= -1(1/2)
= -2
→ slope of tangent is -2
4. Find the equation of the tangent
y - 4 = -2(x - 3)
→ y - 4 = -2x + 6
→ y + 2x = 10
5. Find the point where the tangent crosses the x-axis
x-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 Pythagoras
distance = √(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.
Wiki User
∙ 6y ago
Wiki User
∙ 6y agoThe distance from the x-axis to the centre of the circle is 3 units, irrespective of where a tangent might be.
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What is the length of tangent line from -2 3 to the point of contact with the circle x2 plus y2 plus 6x plus 10y minus 2 equals 0?
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What is the distance from the centre of a circle to a point on the x axis that forms a tangent line when it meets the circle of x2 plus y2 -2x -6y plus 5 equals 0 at the point 3 and 4?
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What is the equation of the tangent line that meets the circle x2 -4x plus y2 -6y equals 4 at the point 6 4?
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What is the point of contact when the tangent line y -3x -5 equals 0 touches the circle x2 plus y2 -2x plus 4y -5 equals 0?
It works out that the tangent line of y -3x -5 = 0 makes contact with the circle x^2 +y^2 -2x +4y -5 = 0 at the coordinate of (-2, -1) on the coordinated grid.
What is the point of contact when the tangent line y -3x -5 equals 0 meets the circle x2 plus y2 -2x plus 4y -5 equals 0?
It works out that the tangent line of y -3x -5 = 0 makes contact with the circle of x^2 + y^2 -2x +4y -5 = 0 at (-2, -1)
What is the distance to the center of the circle x2 plus y2 -2x -6y plus 5 equals 0 from the x axis at the same point when the tangent line meets the x axis from the point 3 4?
Equation of circle: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 = 5 Center of circle: (1, 3) Tangent line from (3, 4) meets the x axis at: (5, 0) Distance from (5, 0) to (1, 3) = 5 using the distance formula
What is the equation for a circle with its center at the origin and a tangent whose equation is y equals 7?
x2 + y2 = 49
What is the distance from a point on the x axis to the centre of the circle x2 plus y2 -2x -6y plus 5 equals 0 from the same point where a tangent line meets the circle at 3 4?
Equation of circle: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x -1)^2 +(y -3)^2 = 5 which is radius squared Center of circle: (1, 3) Tangent line is at right angles to the radius at (3, 4) and meets the x axis at (5, 0) Distance from point (5, 0) to center of circle (1, 3) = 5 units using distance formula
What is the distance from a defined point on the x axis to the centre of circle x2 plus y2 -2x -6y plus 5 equals 0 when its tangent is at 3 4 on the Cartesian plane?
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What is the length of the tangent line from the point 8 2 to a point where it touches the circle of x2 plus y2 -4x -8y -5 equals 0 on the Cartesian plane?
The distance from (8, 2) to the center of the circle forms the hypotenuse of a right angle triangle with the circle's radius meeting the tangent line at right angles and so:- Equation of the circle: x^2 +y^2 -4x -8y -5 = 0 Completing the squares: (x-2)^2 +(y-4)^2 = 25 Center of circle: (2, 4) Radius of circle: 5 Distance from (8, 2) to (2, 4): 2 times square root of 10 Using Pythagoras' theorem: distance squared minus radius squared = 15 Therefore length of the tangent line is the square root of 15
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 length of the tangent line from the point 8 2 to a point where it meets the circle x2 plus y2 -4x -8y -5 equals 0?
Equation of circle: x^2 +y^2 -4x -8y -5 = 0 Completing the squares: (x-2)^2 +(y-4)^2 = 25 which is radius squared Center of circle: (2, 4) Tangent line originates from: (8, 2) Distance from (8, 2) to (2, 4) is sq rt of 40 which is hypotenuse of right angle triangle Using Pythagoras theorem: distance^2 minus radius^2 = 15 Therefore length of tangent line is the square root of 15
What is the length of the tangent line from the point 9 0 to the circle x2 plus 8x plus y2 -9 equals 0?
Equation of circle: x^2 +8x +y^2 -9 = 0 Completing the square: (x+4)^2 +y^2 = 25 Radius of circle: 5 Center of circle: (-4, 0) Distance from (9, 0) to (-4, 0) = 13 which is the hypotenuse of a right angle triangle Using Pythagoras: 13^2 -5^2 = 144 and its square root is 12 Therefore the length of the tangent line is 12 units Note that the tangent line of a circle meets the radius of the circle at right angles
What are the lengths of the tangent lines from the point 7 -2 to the circle x2 plus y2 -10y -24 equals 0 on the Cartesian plane?
Equation of circle: x^2 +y^2 -10y -24 = 0 Completing the square: x^2+(y-5)^2 = 49 Center of circle: (0, 5) Radius of circle: 7 Distance from (7, -2) to (0, 5) = sq rt of 98 and is the hypotenuse of a right triangle Using Pythagoras: theorem: distance^2 minus radius^2 = 49 Therefore lengths of tangent lines are square root of 49 = 7 units
What is the length of the tangent line from the point 9 0 to a point where it touches the circle of x2 plus 8x plus y2 -9 equals 0?
Equation of circle: x^2 +8x +y^2 -9 = 0 Completing the square: (x+4)^2 +y^2 = 25 Radius of circle: 5 Center of circle: (-4, 0) Distance from (9, 0) to (-4, 0) is 13 which is the hypotenuse of a right angle triangle Using Pythagoras' theorem: 13^2 -5^2 = 144 and its square root is 12 Therefore length of tangent line is: 12 units Note that a tangent line always meets the radius of a circle at right angles.
What is the length of the tangent line from the coordinate of 9 0 to a point where it touches the circle of x2 plus 8x plus y2 -9 equals 0?
Circle equation: x^2 +8x +y^2 -9 = 0 Completing the square: (x+4)^2 +y^2 = 25 Center of circle: (-4, 0) Radius of circle: 5 Distance from (-4, 0) to (9, 0) = 13 which will be the hypotenuse of a right triangle Length of tangent line using Pythagoras; theorem: 13^2 -5^2 = 144 Therefore length of tangent line is the square root of 144 = 12 units
What is the solution when y equals 2x plus 1 is a tangent to the circle 5y2 plus 5x2 equals 1?
If y = 2x+1 is a tangent line to the circle 5y^2 +5x^2 = 1 then the point of contact is at (-2/5, 1/5) because it has equal roots
