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Circle equation: x^2 +y^2 -2x -6y +5 = 0

Completing the squares: (x-1)^2 +(y-3)^2 = 5

Centre of circle: (1, 3)

Tangent line meets the x-axis at: (0, 5)

Distance from (0, 5) to (1, 3) = 5 units using the distance formula

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The distance from an unspecified point on the x axis to the centre of the circle is any number greater than or equal to 3.

Q: What is the distance from a point on the x axis to the centre of a circle when a tangent line at the point 3 4 meets the circle of x2 plus y2 -2x -6y plus 5 equals 0?

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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

x² + y² - 10 = 49 → x² + y² = 59 = (x - 0)² + (y - 0)² = (√59)² → circle has centre (0, 0) - the origin - and radius √59 The point (7, -2) has a distance from the centre of the circle of: √((7 - 0)² + (-2 - 0)²) = √(7² + (-2)²) = √(49 + 4) = √53 < √59 Which means that the point is INSIDE the circle and all lines drawn from it to a point on the circumference will NOT be a tangent - the lines will CROSS the circumference, not touch it. Thus there is no solution to the problem as posed. -------------------------------------------------------------------- If the equation for the circle is wrong (which is most likely given as how it was stated) please re-submit your question with the correct equation for the circle.

Point of contact: (21, 8) Equation of circle: x^2 -y^2 -8x -16y -209 = 0 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Centre of circle: (4, 8) and its radius is 17 Slope of radius: 0 Slope of tangent: 0 Tangent equation of the circle: x = 21 meaning that the tangent line is parallel to the y axis and that the radius is parallel to the x axis.

A circle with centre (X, Y) and radius r has an equation of the form: (x - X)² + (y - Y)² = r² Completing the square in x and y for the given equation gives: x² + y² + 4x - 6y + 10 = 0 → (x + (4/2))² - (4/2)² + (y - 6/2) - (6/2)² + 10 = 0 → (x + 2)² +(y - 3)² -2² - 3² + 10 = 0 → (x - -2)² + (y - 3)² = 4 + 9 - 10 = 3 → centre of circle is (-2, 3) and radius is √3 Where a tangent meets a circle it forms a right angle with a radius of the circle. Thus the origin, point of contact of tangent and centre of the circle form a right angled triangle with the hypotenuse the side between the origin and the centre of the circle. Thus Pythagoras can be used to find the length of the hypotenuse and the tangent: tangent² + radius² = hypotenuse² → tangent = √(hypotenuse² - radius²) = √((-2 - 0)² + (3 - 0)² - 3) = √((-2)² + (3)² - 3) = √(4 + 9 - 3) = √10 units ≈ 3.16 units

Circle passing through coordinate: (0, 0) Circle equation: x^2 +6 +y^2 -10 = 0 Completing the squares: (x+3)^2 +(y-5)^2 = 34 Centre of circle: (-3, 5) Slope of radius: -5/3 Slope of tangent: 3/5 Tangent equation: y-0 = 3/5(x-0) => y = 3/5x

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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

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 contact point: (3, 4) Slope of radius: ((3-4)/(1-3) = 1/2 Slope of tangent line: -2 Equation of tangent line: y-4 = -2(x-3) => y = -2x+10 Equation tangent rearranged: 2x+y = 10 When y equals 0 then x = 5 or (5, 0) as a coordinate Distance from (5, 0) to (1, 3) = 5 using the distance formula

Point of contact: (3, 4) Circle equation: x^2 +y^2 -2x -6y+5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 -1 -9 +5 = 0 So: (x-1)^2 +(y-3) = 5 Centre of circle: (1, 3) Slope of radius: (3-4)/(1-3) = 1/2 Slope of tangent: -2 Equation of tangent line: y-4 = -2(x-3) => 2x+y = 10 Tangent line meets the x axis at: (5, 0) Using formula distance from (1, 3) to (5, 0) = 5 units

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

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.

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

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)

x² + y² - 10 = 49 → x² + y² = 59 = (x - 0)² + (y - 0)² = (√59)² → circle has centre (0, 0) - the origin - and radius √59 The point (7, -2) has a distance from the centre of the circle of: √((7 - 0)² + (-2 - 0)²) = √(7² + (-2)²) = √(49 + 4) = √53 < √59 Which means that the point is INSIDE the circle and all lines drawn from it to a point on the circumference will NOT be a tangent - the lines will CROSS the circumference, not touch it. Thus there is no solution to the problem as posed. -------------------------------------------------------------------- If the equation for the circle is wrong (which is most likely given as how it was stated) please re-submit your question with the correct equation for the circle.

Point of contact: (21, 8) Equation of circle: x^2 -y^2 -8x -16y -209 = 0 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Centre of circle: (4, 8) and its radius is 17 Slope of radius: 0 Slope of tangent: 0 Tangent equation of the circle: x = 21 meaning that the tangent line is parallel to the y axis and that the radius is parallel to the x axis.

A circle with centre (X, Y) and radius r has an equation of the form: (x - X)² + (y - Y)² = r² Completing the square in x and y for the given equation gives: x² + y² + 4x - 6y + 10 = 0 → (x + (4/2))² - (4/2)² + (y - 6/2) - (6/2)² + 10 = 0 → (x + 2)² +(y - 3)² -2² - 3² + 10 = 0 → (x - -2)² + (y - 3)² = 4 + 9 - 10 = 3 → centre of circle is (-2, 3) and radius is √3 Where a tangent meets a circle it forms a right angle with a radius of the circle. Thus the origin, point of contact of tangent and centre of the circle form a right angled triangle with the hypotenuse the side between the origin and the centre of the circle. Thus Pythagoras can be used to find the length of the hypotenuse and the tangent: tangent² + radius² = hypotenuse² → tangent = √(hypotenuse² - radius²) = √((-2 - 0)² + (3 - 0)² - 3) = √((-2)² + (3)² - 3) = √(4 + 9 - 3) = √10 units ≈ 3.16 units

The chord that passes through the center of a circle is called the diameter. The measure of this chord is also called the diameter, and is used in calculations such as finding the circumference of a circle (Circumference equals pi times the diameter).^^wrong...it is eitherSecant or tangent im pretty sure its SECANT though...

Circle passing through coordinate: (0, 0) Circle equation: x^2 +6 +y^2 -10 = 0 Completing the squares: (x+3)^2 +(y-5)^2 = 34 Centre of circle: (-3, 5) Slope of radius: -5/3 Slope of tangent: 3/5 Tangent equation: y-0 = 3/5(x-0) => y = 3/5x