The points A,B,C are(-2,-4),(3,1)and (-2,0). Find the equation of the line passing through A,B,C and?

show that the tangent at B is parallel to diameter through C.


Please show workings.

The points A,B,C are(-2,-4),(3,1)and (-2,0). Find the equation of the line passing through A,B,C and?
i'm assuming you mean that the line through A, B, C is a circle.





A (-2,-4)


B (3,1)


C (-2,0)





to get the center of circle:


we get the midpoint of AC and BC


midpoint of AC:


( (-2 + -2)/2 , (-4 + 0)/2 ) -%26gt; (-2, -2)


midpoint of BC:


( (-2 + 3)/2 , (1 + 0)/2 ) -%26gt; (1/2, 1/2)


we get a line perpendicular to AC through its midpoint


since AC is a vertical line (with equation x = -2), the perpendicular line through its midpoint is:


y = -2


similarly, we get a line perpendicular to BC through its midpoint


getting the equation of BC:


1 = 3m + b


0 = -2m + b


subtracting the 2nd equation from the 1st,


1 = 5m


m = 1/5, b = 2/5


equation of BC: y = (1/5)x + 2/5


so the perpendicular through BC has slope -5


hence, 1/2 = -5/2 + b


b = 3


the eqn is y = -5x + 3





we get the intersection of the 2 perpendicular lines to get the center of the circle:


y = -2 = -5x + 3


-5 = -5x


x = 1, y = -2


hence, the center of the circle is (1, -2)


to get the radius of the circle, we get the distance of any of the three points from the center


using the distance formula, the distance of C from center is:


d = sqrt [ (1 + 2)^2 + (-2 - 0)^2 ] = sqrt (13)


hence, the equation of the circle is


(x - 1)^2 + (y + 2)^2 = 13





getting the slope of the diameter through C:


-2 = m + b


0 = -2m + b


subtracting the 2nd equation from the first,


-2 = 3m


m = -2/3





getting the slope of the line through B and the center,


-2 = m + b


1 = 3m + b


subtracting the 2nd from the 1st,


-3 = -2m


m = 3/2





getting the slope of the tangent through B, (the slope is the negative reciprocal of the slope of the line connecting B and the center)


m = -2/3





since the slopes of the tangent through B and of the diameter through C is the same, these 2 are parallel.
Reply:It seems that you have made an error in typing your question. You want the equation of the circle and not line.


The equation of a circle having A(-2, -4) and C(-2, 0) as endpoints of a diameter is


(x+2)(x+2) + (y + 4)(y - 0) = 0


=%26gt; x^2 + y^2 + 4x + 4y + 4 = 0


and the equation of the line through A and C is x = -2


=%26gt; eqn. of any circle through the points of intersection of the above circle and line is


x^2 + y^2 + 4x + 4y + 4 + k(x + 2) = 0


If B(3, 1) lies on it,


3^2 + 1^2 + 4*3 + 4*1 + 4 + k(3 + 2) = 0 =%26gt; k = -30/5


=%26gt; eqn. of the required circle is


x^2 + y^2 + 4x + 4y + 4 - (30/5)(x + 2) = 0


=%26gt; 5x^2 + 5y^2 - 10x + 20y - 40 = 0


=%26gt; x^2 + y^2 - 2x + 4y - 8 = 0





The eqn of tangent at B (3, 1) is


x*3 + y*1 -(x + 3) + 2(y + 1) - 8 = 0


=%26gt; 2x + 3y - 9 = 0





Center of the circle is O(1, -2)


Slope of line through O and C = (0 + 2) / (-2 - 1) = --2/3





As the slope of diameter through C is the same as the slope of the tangent at B, they are parallel.
Reply:from any three lines there must be pass one and only one circle .


The equation of circle will be


x^2+y^2+2gx+2fy+d=0


(-2,-4),(3,1),(-2,0) will lie on above equation,


so, above equation will be,


x^2+y^2-2x+4y-8=0 ,


hence center is (1,-2),


tangent to the circle at point (3,1) will be


2 x+3y-9=0____(1)


now equation of line through c(-2,0) and center(1,-2) will be


2x+3y+4=0_______(2) ,


from hear (1)%26amp;(2) are parallal.
Reply:July 02, 2007


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i'm assuming you mean that the line through A, B, C is a circle.





A (-2,-4)


B (3,1)


C (-2,0)





to get the center of circle:


we get the midpoint of AC and BC


midpoint of AC:


( (-2 + -2)/2 , (-4 + 0)/2 ) -%26gt; (-2, -2)


midpoint of BC:


( (-2 + 3)/2 , (1 + 0)/2 ) -%26gt; (1/2, 1/2)


we get a line perpendicular to AC through its midpoint


since AC is a vertical line (with equation x = -2), the perpendicular line through its midpoint is:


y = -2


similarly, we get a line perpendicular to BC through its midpoint


getting the equation of BC:


1 = 3m + b


0 = -2m + b


subtracting the 2nd equation from the 1st,


1 = 5m


m = 1/5, b = 2/5


equation of BC: y = (1/5)x + 2/5


so the perpendicular through BC has slope -5


hence, 1/2 = -5/2 + b


b = 3


the eqn is y = -5x + 3





we get the intersection of the 2 perpendicular lines to get the center of the circle:


y = -2 = -5x + 3


-5 = -5x


x = 1, y = -2


hence, the center of the circle is (1, -2)


to get the radius of the circle, we get the distance of any of the three points from the center


using the distance formula, the distance of C from center is:


d = sqrt [ (1 + 2)^2 + (-2 - 0)^2 ] = sqrt (13)


hence, the equation of the circle is


(x - 1)^2 + (y + 2)^2 = 13





getting the slope of the diameter through C:


-2 = m + b


0 = -2m + b


subtracting the 2nd equation from the first,


-2 = 3m


m = -2/3





getting the slope of the line through B and the center,


-2 = m + b


1 = 3m + b


subtracting the 2nd from the 1st,


-3 = -2m


m = 3/2





getting the slope of the tangent through B, (the slope is the negative reciprocal of the slope of the line connecting B and the center)


m = -2/3





since the slopes of the tangent through B and of the diameter through C is the same, these 2 are parallel.

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