Question

If the center of Circle with equation x²+y²-ax-by-12=0 is (2,3) find the value of a and b

Answer 1

Answer:

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Answer 2

Answer:

The values of $a=4$ and $b=6$.

Step-by-step explanation:

Step 1 of 3

Consider the general equation of the circle as follows:

$(x-h)^2+(y-k)^2=r^2$ . . . . . (1)

where $(h,k)$ is the center of the circle and $r$ is the radius.

Now,

Expand the equation (1) using the identity, $(a-b)^2=a^2+b^2-2ab$.

$x^2+h^2-2xh+y^2+k^2-2yk=r^2$

⇒ $x^2+y^2-2xh-2yk+h^2+k^2=r^2$ . . . . . (2)

Step 2 of 3

Given: The equation of circle is $x^2+y^2-ax-by-12=0$,

or, $x^2+y^2-ax-by=12$ . . . . . (3)

Also, the center of circle is $(2,3)$, i.e.,

$(h,k)=(2,3)$

⇒ $h=2$ and $k=3$

Substitute the values of $h$ and $k$ in the equation (2), we get

$x^2+y^2-2x(2)-2y(3)+(2)^2+(3)^2=r^2$

$x^2+y^2-4x-6y+4+9=r^2$

$x^2+y^2-4x-6y=r^2-13$ . . . . . (4)

Step 3 of 3

Compare the equations (3) and (4) as follows:

⇒ $a=4$

⇒ $b=6$

⇒ $r^2-13=12$

⇒ $r^2=12+13$

⇒ $r^2=25$ or $r=5$ (Radius)

Therefore, the value of $a$ is $4$ and the value of $b$ is $6$.

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