![]() ⇒ x 2 + y 2 + 3 x + y - 6 = 0 which is the required equation of the circle. Įquation of the circle with end points of the diameter as( x 1 , y 1 ) and ( x 2 , y 2 ) given in įind the general equation of the circle whose diameter is the line segment joining the points (−4, −2) and (1,1). That is, = −1 yielding the equation of the required circle as Therefore, the product of slopes of AP and PB is equal to -1. Then Ð APB = π/2 (angle in a semi-circle). Let A( x 1 , y 1 ) and B( x 2 , y 2 ) be the two extremities of the diameter AB , and P( x, y) be any point on the circle. The equation of a circle with ( x 1 , y 1 ) and ( x 2 , y 2 ) as extremities of one of the diameters of the circle is ( x − x 1 )( x − x 2 ) + ( y − y 1)( y − y 2) = 0. ![]() So the line x + y -1 = 0 passes through the centre and therefore the line x + y -1 = 0 is a diameter of the circle for all possible values of c . ![]() Therefore, the equation of the required circle is x 2 + y 2 + 3 x + y -11 = 0.ĭetermine whether x + y −1 = 0 is the equation of a diameter of the circle x 2 + y 2 − 6 x + 4 y + c = 0 for all possible values of c .Ĭentre of the circle is (3, -2) which lies on x + y -1 = 0. The chord 3 x + y + 5 = 0 is a diameter of this circle if the centre lies on the chord. ⇒ ( x - ( - 3 ) ) 2 + ( y - ( - 4 ) ) 2 = 3 2įind the equation of the circle described on the chord 3 x + y + 5 = 0 of the circle x 2 + y 2 = 16 as diameter.Įquation of the circle passing through the points of intersection of the chord and circle by Theorem 5.1 is x 2 + y 2 -16 + l (3 x + y + 5) = 0. Find the general equation of a circle with centre (−3, −4) and radius 3 units.Įquation of the circle in standard form is ( x - h ) 2 + ( y - k ) 2 = r 2
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