Equation of circle: x^2 +y^2 +6x +10y -2 = 0
Completing the squares: (x+3)^2 +(y+5)^2 = 36
Radius of circle: 6
Center of circle: (-3, -5)
Distance from (-2, 3) to (-3, -5) is sq rt of 65 which is hypotenuse of a right triangle
Using Pythagoras' theorem: square root of 65^2 -6^2 = 29
Therefore length of tangent line is the square root of 29
Note that the tangent line of any circle always meets its radius at right angles which is 90 degrees.
the length of thr direct common tangent will be 2*{1/2 power of (r1*r2)} the answer will be 8 units in this case...
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tan A = (sin A) / (cos A) tan (A)= opposite side length/adjacent side length A is an angle measurement; amount of degrees or radians. If a line is tangent to a curve, it only touches the curve at one point. looks like )| but the line is touching the curve. In a circle, the tangent line touches the circle at one point and is perpinducular to the circle's radius if it is touching that same point.
Its Tangent, APEX "The tangent of an angle is the ratio of the opposite leg length to the adjacent leg length."
circumfrence off the circle
the length of thr direct common tangent will be 2*{1/2 power of (r1*r2)} the answer will be 8 units in this case...
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The tangent of a circle is perpendicular to the radius to the point of contact (Xc, Yc).The point (-2, 3), the centre of the circle (Xo, Yo) and the point of contact of the tangent (Xc, Yc) form a right angle triangle.The leg from the point (-2, 3) to the point of contact (Xc, Yc) is the required lengthThe leg from the centre of the circle (Xo, Yo) to the point of contact (Xc, Yc) has length equal to the radius (r) of the circleThe hypotenuse is the length between the point (-2, 3) and the centre of the circle (Xo, Yo).To solve this:Find the centre (Xo, Yo) of the circle, and its radius r.Use Pythagoras to find the length between the point (-2, 3) and the centre of the circle (Xo, Yo)Use Pythagoras to find the length between the point (-2, 3) and the point of contact (Xc, Yc) of the tangent - the required length.Hint: a circle with centre (Xo, Yo) and radius r has an equation of the form:(x - Xo)² + (y - Yo)² = r²Have a go at solving it now you know how, before reading the solution below:------------------------------------------------------------------------------Circle:x² + y² + 6x + 10y - 2 = 0→ x² + 6x + y² + 10y - 2 = 0→ (x + (6/2))² - (6/2)² + (y + (10/2))² - (10/2)² - 2 = 0→ (x + 3)² - 9 + (y + 5)² - 25 - 2 = 0→ (x + 3)² + (y + 5)² = 36 = 6²→ Circle has centre (-3, -5) and radius 6Line from centre of circle (-3, -5) to the given point (-2, 3):Using Pythagoras to find length of a line between two points (x1, y1) and (x2, y2):length = √((x2 - x1)² + (y2 - y1)²)To find length between given point (-2, 3) and centre of circle (-3, -5)→ length = √((-5 - -2)² + (-3 - -3)²)= √((-3)² + (-6)²)= √45Tangent line segment:Using Pythagoras to find length of tangent between point (-2, 3) and its point of contact with the circle:centre_to_point² = tangent² + radius²→ tangent = √(centre_to_point² - radius²)= √((√45)² + 6²)= √(45 + 36)= √81= 9The length is 9 units.
The tangent of a circle is perpendicular to the radius to the point of contact (Xc, Yc).The point (0, 0), the centre of the circle (Xo, Yo) and the point of contact of the tangent (Xc, Yc) form a right angle triangle.The leg from the point (0, 0) to the point of contact (Xc, Yc) is the required lengthThe leg from the centre of the circle (Xo, Yo) to the point of contact (Xc, Yc) has length equal to the radius (r) of the circleThe hypotenuse is the length between the point (0, 0) and the centre of the circle (Xo, Yo).To solve this:Find the centre (Xo, Yo) of the circle, and its radius r.Use Pythagoras to find the length between the point (0, 0) and the centre of the circle (Xo, Yo)Use Pythagoras to find the length between the point (0, 0) and the point of contact (Xc, Yc) of the tangent - the required length.Hint: a circle with centre (Xo, Yo) and radius r has an equation of the form:(x - Xo)² + (y - Yo)² = r²Have a go at solving it now you know how, before reading the solution below:------------------------------------------------------------------------------Circle:x² + y² + 4x - 6y + 10 = 0→ x² + 4x + y² - 6y + 10 = 0→ (x + (4/2))² - (4/2)² + (y + (-6/2))² - (-6/2)² +10 = 0→ (x + 2)² - 4 + (y - 3)² - 9 + 10 = 0→ (x + 2)² + (y - 3)² = 3 = radius²→ Circle has centre (-2, 3) and radius √3Line from centre of circle (-2, 3) to the given point (0, 0):Using Pythagoras to find length of a line between two points (x1, y1) and (x2, y2):length = √((x2 - x1)² + (y2 - y1)²)To find length between given point (0, 0) and centre of circle (-2, 3)→ length = √((0 - -2)² + (0 - 3)²)= √(2² + (-3)²)= √13Tangent line segment:Using Pythagoras to find length of tangent between point (0, 0) and its point of contact with the circle:centre_to_point² = tangent² + radius²→ tangent = √(centre_to_point² - radius²)= √((√13)² + (√3)²)= √(13 + 3)= √16= 4
The tangent of a circle is perpendicular to the radius to the point of contact (Xc, Yc).The point (Xg, Yg), the centre of the circle (Xo, Yo) and the point of contact of the tangent (Xc, Yc) form a right angle triangle.The leg from the point (Xg, Yg) to the point of contact (Xc, Yc) is the required lengthThe leg from the centre of the circle (Xo, Yo) to the point of contact (Xc, Yc) has length equal to the radius (r) of the circleThe hypotenuse is the length between the point (0, 0) and the centre of the circle (Xo, Yo).To solve this:Find the centre (Xo, Yo) of the circle, and its radius r;Use Pythagoras to find the length between the point (Xg, Yg) and the centre of the circle (Xo, Yo);Use Pythagoras to find the length between the point (Xg, Yg) and the point of contact (Xc, Yc) of the tangent - the required length.Hint: a circle with centre (Xo, Yo) and radius r has an equation of the form:(x - Xo)² + (y - Yo)² = r²Have a go at solving it now you know how, before reading the solution below:------------------------------------------------------------------------------Circle:x² + y² - 4x - 8y - 5 = 0→ x² - 4x + y² - 8y - 5 = 0→ (x - (4/2))² - (4/2)² + (y - (8/2))² - (8/2)² - 5= 0→ (x - 2)² - 4 + (y - 4)² - 16 - 5 = 0→ (x - 2)² + (y - 4)² = 25 = radius²→ Circle has centre (2, 4) and radius √25 = 5Line from centre of circle (2, 4) to the given point (8, 2):Using Pythagoras to find length of a line between two points (x1, y1) and (x2, y2):length = √((x2 - x1)² + (y2 - y1)²)To find length between given point (8, 2) and centre of circle (2, 4)→ length = √((2 - 8)² + (4 - 2)²)= √((-6)² + 2²)= √40Tangent line segment:Using Pythagoras to find length of tangent between point (8, 2) and its point of contact with the circle:centre_to_point² = tangent² + radius²→ tangent = √(centre_to_point² - radius²)= √((√40)² + 25)= √65≈ 8.06
*If two pair of tangent of inner circle making angles on the circumference of outer circle then the angles so formed are equal . *Any two tangent of inner circle within the outer circle's circumference are equal in length .
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
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
tan A = (sin A) / (cos A) tan (A)= opposite side length/adjacent side length A is an angle measurement; amount of degrees or radians. If a line is tangent to a curve, it only touches the curve at one point. looks like )| but the line is touching the curve. In a circle, the tangent line touches the circle at one point and is perpinducular to the circle's radius if it is touching that same point.
The circles are concentric with centre (0,0). The radius of the outer circle is sqrt(72), that of the inner circle is sqrt(18). By Pythagoras, the length of the semichord is sqrt(72 - 18) = sqrt(54) units. Therefore the chord is 2*sqrt(54) = 6*sqrt(6) = 14.679 units (approx).
Course Hero Question A tangent segment and a secant segment are drawn to a circle from a point outside the circle. The length of the tangent segment is 15 inches. The... Answer · 0 votes Length of interior part of secant = 40 inches Please see attached image for diagram with work shown Image transcriptions The tangent—secant theorem states that if a tangent and a secant are drawn from the same external point, the length of the tangent squared is equal to the external part of the secant multiplied by the whole segment. 15_ Let x = the length of the inner segment of the secant II'I ' Length whole secant = length interior of secant + length exterior of secant 5 in = x + 5 (tangent? = (length exterior) * (length whole secant) (15)2 = (5) * (x + 5) 225 = 5x + 25 200: 5x 40:): Measure of internal segment = 40 inches More
It is the same length from the centre to any point on the circumference so just measure it