Question

What is the value of a for the following circle in general form? p+y+ar+by+c=0 C on 6. on Answer here

Answer 1

The general equation of a circle with centre (h, k) express as;

`(x-h)^2+(y-k)^2=r^2`

The given circle given as;

with centre define as; (-4, 2) and one passing point (x, y) = (-2, 2)

Substitute the center as h = -4 and k = 2 and one passing point as x = -2 and y = 2in the general equation of circle;

`\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (-2-(-4))^2+(2-2)^2=r^2 \\ (-2+4)^2+0=r^2 \\ r^2=2^2 \\ r=2 \end{gathered}`

Now substitute the center (-4, 2) and radius r =2 in the general equation of circle and simplify;

`\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (x-(-4))^2+(y-2)^2=2^2 \\ (x+4)^2+(y-2)^2=4 \\ x^2+16+8x+y^2+4-4y=4 \\ x^2+y^2+8x-4y+16=0 \end{gathered}`

On comparing the above equation with the given expression we get;

`\begin{gathered} x^2+y^2+8x-4y+16=x^2+y^2+ax+by+c \\ a\text{ =8, b = -4 and c = 16} \end{gathered}`

Answer : a = 8

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