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What are the equations of the tangents to the circle x2 plus y2 -6x plus 4y plus 5 equals 0 at the points where they meet the x axis on the Cartesian plane?
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∙ 6y ago
Circle equation: x^2 +y^2 -6x +4y +5 = 0
Completing the squares: (x-3)^2 +(y+2)^2 = 8
Center of circle: (3, -2)
Radius of circle: square root of 8
The radius of the circle will touch the points of (1, 0) and (5, 0) on the x axis
The tangent slope at (1, 0) is 1
The tangent slope at (5, 0) is -1
Equations of the tangents are: y = x-1 and y = -x+5
Wiki User
∙ 6y ago
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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.
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What is the relationship between the line y equals x plus 4 and the circle x squared plus y squared -8x plus 4y equals 30 on the Cartesian plane explaining how and why?
Equations: y = x+4 and x^2 +y^2 -8x +4y = 30 The given equations will finally form a quadratic equation such as: x^2 +2x +1 = 0 Discriminant: 2^2 -4*(1*1) = 0 meaning there are equal roots Because the discriminant has equal roots the line is a tangent to the circle In fact the line makes contact with the circle at (-1, 3) on the Cartesian plane
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.
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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.
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What is y equals 2x-1 and y equals 2x plus 1?
They are two equations in two unknown variables (x and y), which are inconsistent. That is to say, there is no simultaneous solution. An alternative approach is to say that they are the equations of two lines in the Cartesian plane. The lines are parallel and so they do not meet indicating that there is no simultaneous solution.
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