SOLUTION: how can i find the equation (in standard form) of a circle that is tangent to the y-axiz with the center (-8,-7)?

Question 29588: how can i find the equation (in standard form) of a circle that is tangent to the y-axiz with the center (-8,-7)?
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
how can i find the equation (in standard form) of a circle that is tangent to the y-axiz with the center (-8,-7)?
The centre of the given circle is C(-8,-7) ( a point in the third quadrant)
and by data the y-axis is tangent to the circle say at B.
(Please draw the figure- I am unable to do it in this answer box)
Then CB = radius of the circle = 8 units
(the modulus of the x-coordinate of the center
as y-axis is tangent to the circle)
The equation to the circle with center C(-8,-7) and radius = 8 is given by
[x-(-8)]^2+[y-(-7)^2]= 8^2
(using formula (x-h)^2+(y-k)^2 = r^2 which is the general form of equation to a circle with centre at (h,k) and radius =r )
That is (x+8)^2 +(y+7)^2 = 8^2
On expansion
[x^2+8^2+16x]+[y^2+7^2+14y]=8^2 (using (a+b)^2 = a^2 +b^2 +2ab )
Grouping like terms
x^2+y^2+16x+14y+49 =0
which is the required equation in the standard form
Verification:center= [(-1/2 of coefficient of x),(-1/2 of coefficient of y)
when the circle is in the standard form
=[(-1/2)X(16),(-1/2)X(14)] = (-8,-7) which is correct
radius = sqrt[(-8)^2+(-7)^2-(49)]= sqrt(8^2) = 8 which is correct
using formula r = sqrt[g^2+f^2-c] where (-g,-f)=(h,k) is the center and c is the free constant in the standard form
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