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Given that the circle A whose equation is given by x^2+y^2+4x+6y-12=0 passes through...

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user5451174 | eNotes Newbie

Posted July 28, 2013 at 5:17 AM via web

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Given that the circle A whose equation is given by x^2+y^2+4x+6y-12=0 passes through the point p(-5,-7) find, 

i)the centre of circle and radius 

ii) the coordinates of the other end point of the diameter through P

iii) the equation of circle which shares the same centre as circle A and touches the x-axis. 

Answer I) r=5 

           II)(1,1)

           III)x^2+y^2+4x+6y+4=0

i do not know how to do part 3 please help 

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aruv | High School Teacher | Valedictorian

Posted July 28, 2013 at 5:44 AM (Answer #1)

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Equation of the circle is

`x^2+y^2+4x+6y-12=0`

`x^2+4x+4+y^2+6y+9=12+4+9`

`(x+2)^2+(y+3)^2=(5)^2`

(i)Thus centre of the circle = (-2,-3)

radius of the circle= r

(ii) Let the other end point of diameter opposite to (-5,-7) be `_(x_1,y_1)`

Then

`-2=(x_1-5)/2 => -4=x_1-5`

`x_1=1`

`and`

`-3=(y_1-7)/2=-6=y_1-7`

`y_1=-6+7=1`

`y_1=1`

`Thus`

`(x_1,y_1)=(1,1)`

(iii) Equation of x-axis is y=0

length of perpendicular from (-2,-3) to x-axis=-3 

`(x+2)^2+(y+3)^2=(-3)^2`

`x^2+4x+4+y^2+6y+9=9`

`x^2+y^2+4x+6+4=0`

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