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...
Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares (x -3)^2 +(y +2)^2 = 8 Centre of circle: (3, -2) Radius of circle: square root of 8 Points of contact are at: (1, 0) and (5, 0) where the radii touches the x axis Slope of 1st tangent line: 1 Slope of 2nd tangent line: -1 Equation of 1st tangent: y -0 = 1(x -1) => y = x -1 Equation of 2nd tangent: y -0 = -1(x -5) => y = -x +5
2 circles that intersect each other
arc
First find the slope of the circle's radius as follows:- Equation of circle: x^2 +10x +y^2 -2y -39 = 0 Completing the squares: (x+5)^2 + (y-1)^2 -25 -1 -39 = 0 So: (x+5)^2 +(y-1)^2 = 65 Centre of circle: (-5, 1) and point of contact (3, 2) Slope of radius: (1-2)/(-5-3) = 1/8 which is perpendicular to the tangent line Slope of tangent line: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26 Tangent equation in its general form: 8x+y-26 = 0
A line can be tangent to a circle in which case it intersects it in one point, it can intersect it in two points, or no points at all. So the choices are 0,1 or 2.
No, a circle can never pass through three points of a straight line. The circle will touch 1) no points of the line, 2) one point of the line (which is now tangent to the circle), or 3) two points of the line. A line can contain (at most) twopoints that lie on the line.
It depends what kind of curve you're talking about. if it's a circle, and the line is tangent to the circle, then one. If it's a circle, and the line is not tangent to the circle, then two. But if it's a goofy shaped curve then it could be any number. But the most likely answer to your question is 2.
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...
Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares (x -3)^2 +(y +2)^2 = 8 Centre of circle: (3, -2) Radius of circle: square root of 8 Points of contact are at: (1, 0) and (5, 0) where the radii touches the x axis Slope of 1st tangent line: 1 Slope of 2nd tangent line: -1 Equation of 1st tangent: y -0 = 1(x -1) => y = x -1 Equation of 2nd tangent: y -0 = -1(x -5) => y = -x +5
180/pi=57.2958 or two lines intersect in one point. that point is the center of a circle. if the angle between the two lines is 180/pi (57.2958...) then the radius of the circle is equal to the length of the section of circumference between the points where the 2 lines intersect the circle. This is true whatever the radius of the circle is.
There may be an easier way, but this is one way to do it: 1) establish the centerpoint of the given circle. - pick two random points on the circle and draw intersecting arcs A1 A2 of equal radius centered on those points. Then draw a line through the two points where A1 and A2 intersect. This line will pass through the circle center. - repeat with two other points. You now have two lines that intersect at the circle centerpoint C. 2) draw a line perpendicular to the given line that passes through C. - draw an arc centered on C that intersects the given line twice. Repeat the bisecting procedure as before using those two intersection points. Call the newly created line L. 3) draw the desired tangent line. - call the point where L intersects the given circle P. (Note that there are actually two such points, since there are two solutions to your problem - one on the near side and one on the far side.) Generate two equidistant points on L by drawing a small circle centered on P. Use those new points for the old bisection procedure and you have your answer!
Step I: Show that both points are outside the smaller circles. Possibly by showing that distance from each point to the centre of the circle is greater than its radius. Step 2: Show that the line between the two points touches the circle at exactly one point. This would be by simultaneous solution of the equations of the line and the circle.
Zero; parallel lines never intersect.
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
They need not intersect at all, but if they do, it will be along a straight line and so comprise infinitely many points.
2 circles that intersect each other