Since the circle has intercept ‘a’ from x – axis, the circle must pass through (a, 0) and (-a, 0) as it already passes through the origin. Also,since the circle has intercept ‘b’ from x – axis, the...
Find the equation of the circle passing through the origin and the points where the line 3x + 4y = 12 meets the axes of coordinates.
The line 3x + 4y = 12 The value of x is 0 on meeting the y – axis. So, \[\begin{array}{*{35}{l}} 3\left( 0 \right)\text{ }+\text{ }4y\text{ }=\text{ }12 \\ 4y\text{ }=\text{ }12 \\ y\text{...
Find the equation of the circle circumscribing the rectangle whose sides are x – 3y = 4, 3x + y = 22, x – 3y = 14 and 3x + y = 62.
The sides \[\begin{array}{*{35}{l}} x\text{ }-\text{ }3y\text{ }=\text{ }4\text{ }\ldots .\text{ }\left( 1 \right) \\ 3x\text{ }+\text{ }y\text{ }=\text{ }22\text{ }\ldots \text{ }\left( 2 \right) ...
The sides of a squares are x = 6, x = 9, y = 3 and y = 6. Find the equation of a circle drawn on the diagonal of the square as its diameter.
The sides of a squares are x = 6, x = 9, y = 3 and y = 6. assuming A, B, C, D be the vertices of the square. we get, the coordinates as: A = (6, 3) B = (9, 3) C = (9, 6) D = (6, 6) the equation of...
Find the equation of the circle the end points of whose diameter are the centres of the circles x^2 + y^2 + 6x – 14y – 1 = 0 and x^2 + y^2 – 4x + 10y – 2 = 0.
x2 + y2 + 6x – 14y – 1 = 0…. (1) So the centre \[\begin{array}{*{35}{l}} =\text{ }\left[ \left( -6/2 \right),\text{ }-\left( -14/2 \right) \right] \\ =\text{ }\left[ -3,\text{ }7 \right] \\...
Find the equation of the circle, the end points of whose diameter are (2, -3) and (-2, 4). Find its centre and radius.
The diameters (2, -3) and (-2, 4). By using the formula, Centre = (-a, -b) \[\begin{array}{*{35}{l}} =\text{ }\left[ \left( 2-2 \right)/2,\text{ }\left( -3+4 \right)/2 \right] \\ =\text{ }\left(...