Find the equation of the circle if the chord of the circle joining (1,
2) and `(-3,1)`
subtents `90^0`
at the center of the circle.
Text Solution
Verified by Experts
The chord joining points A(1,2) and B(-3,1) subtends `90^(@)` at the center of the circle. Then AB subtend an angle `45^(@)` at point C on the circumference. Therefore, the equation of the circle is `(x-1)(x+3)+(y-2)(y-1)` `= +-cot 45^(@)[(y-2)(x+3)-(x-1)(y-1)]` or `x^(2)+y^(2)+2x-3y-1= +-[4y-x-7]` or `x^(2)+y^(2)+3x-7y+6=0` and `x^(2)+y^(2)+x+y-8=0`
Find the equation of the chord of the circle x^2+y^2=9 whose middle point is (1,-2)
Find the equation of the circle which touches the x-axis and whose center is (1, 2).
Show that the points (2, 0), (5, -3), (2, -6) and (-1, -3) are concyclic, find the equation of the circle on which the points lie and the coordinates of the centre of the circle.
Find the equation of the circle having center at (1,2) and which touches x+y=-1 .
Find the equation of each of the circles passing through the points : (0, 0), (1, 2), (2, 0)
Find the equation of each of the circles passing through the points : (2, -1), (2, 3), (4, -1)
Find the equation of the circle which passes through the points (3,-2)a n d(-2,0) and the center lies on the line 2x-y=3
Find the equation of the circle whose diameter is the line joining the points (-4, 3) and (12, -1). Find also the intercept made by it on y-axis
Find the equation of the circle passing through the points (2,3) and (-1,-1) and whose centre is on the line x-3y-11=0.
Find the locus of the midpoint of the chord of the circle x^2+y^2-2x-2y=0 , which makes an angle of 120^0 at the center.