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What is the point of contact between the line y equals x plus 4 and the circle x2 plus y2 -8x plus 4y equals 30?
Wiki User
∙ 6y ago
Equations: y = x+4 and x^2 +y^2 -8x +4y = 30
It appears that the given line is a tangent line to the given circle and the point of contact works out as (-1, 3)
Wiki User
∙ 6y ago
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The angle between a tangent and the radius of a circle att the point of contact is?
It is a right angle
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
What is the length of the tangent line from the point -2 3 to a point where it touches the circle of x2 plus y2 plus 6x plus 10y -2 equals 0?
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.
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?
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
Triangle pqr is an isosceles triangle in which pq equals pr equals 5cm it is circumscribed about a circle prove that qr is bisected at point of contact?
Suppose the circle meets QR at A, RP at B and PQ at C. PQ = PR (given) so PC + CQ = PB + BR. But PC and PB are tangents to the circle from point P, so PC = PB. Therefore CQ = BR Now CQ and AQ are tangents to the circle from point Q, so CQ = AQ and BR and AR are tangents to the circle from point R, so BR = AR Therefore AQ = AR, that is, A is the midpoint of QR.
The angle between a tangent and the radius of a circle att the point of contact is?
It is a right angle
What is the distance between the center of a circle and a point on the circle?
The radius is the distance between the center of a circle and a point on the circle
What is the place where a tangent line intersects a circle?
Definition: a tangent is a line that intersects a circle at exactly one point, the point of intersection is the point of contact or the point of tangency. a tangent is a line that intersects a circle at exactly one point, the point of intersection is the (point of contact) or the **point of tangency**.
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
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 length of the tangent line from the point -2 3 to a point where it touches the circle of x2 plus y2 plus 6x plus 10y -2 equals 0?
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.
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?
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
Triangle pqr is an isosceles triangle in which pq equals pr equals 5cm it is circumscribed about a circle prove that qr is bisected at point of contact?
Suppose the circle meets QR at A, RP at B and PQ at C. PQ = PR (given) so PC + CQ = PB + BR. But PC and PB are tangents to the circle from point P, so PC = PB. Therefore CQ = BR Now CQ and AQ are tangents to the circle from point Q, so CQ = AQ and BR and AR are tangents to the circle from point R, so BR = AR Therefore AQ = AR, that is, A is the midpoint of QR.
What is the length of the tangent line from the origin to a point where it touches the circle x2 plus y2 plus 4x minus 6y plus 10 equals 0 on the Cartesian plane?
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
Point of intersection between a tangent and circle?
A line tyhat's tangent to a circle intersects the circle in exactly one single point. The radius drawn to that point is perpendicular to the tangent.
Ab and ac are tangents to a circle with centre o and radius 5cmif ac equals 12cmfind the perimeter of quadrilateral aboc?
Since AB and AC are tangent to the circle O, it seems that they both are drawn from the same outside point A. As tangents to a circle from an outside point are congruent, AB ≅ BC. Also, a tangent is perpendicular to radius drawn to point of contact. So that OB and OC are congruent radii. Therefore, the perimeter of the quadrilateral ABOC equals to P = 2(12 cm) + 2(5 cm) = 34 cm.
