Question

find eq of circle passing through the point (1,1) and concentric with x2+y2-6x-4y-12=0​

Answer 1

Answer:

It is given that,

Equation of the circle is

x²+y²-6x-4y-12=0

C=(3,2)

And it is passing through the point A(1,1)

AC=√5units

The concentric circle is,

(x-3)²+(y-2)²=5

Step-by-step explanation:

Hope it helps you......

Answer 2

The equation of the circle is

$\boxed{x^2+y^2-6x-4y+8=0}$

Step-by-step explanation:

The given circle is

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

$\implies x^2+y^2-6x-4y=12$

$\implies x^2+y^2-2\times x\times 3+3^2+y^2-2\times y\times 2+2^2-3^2-2^2=12$

$\implies (x-3)^2+(y-2)^2-9-4=12$

$\implies (x-3)^2+(y-2)^2=25$

$\implies (x-3)^2+(y-2)^2=5^2$

Comparing this with general equation of circle $(x-h)^2+(y-k)^2=r^2$ whose centre is $(h,k)$ and radius $r$

We get the centre of the given circle  as $(3,2)$

The radius of the new circle will be distance between $(3,2)$ and $(1,1)$

$\implies r=\sqrt{(3-1)^2+(2-1)^2}$

$\implies r=\sqrt{4+1}$

$\implies r=\sqrt{5}$

Therefore, the equation of the circle

$(x-3)^2+(y-2)^2=(\sqrt{5})^2$

$\implies x^2-6x+9+y^2-4y+4=5$

$\implies x^2+y^2-6x-4y+8=0$

This is the equation of the circle.

Hope this answer is helpful.

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