Question

Find the equation of the circle passes thriugh (2,3) and concentric with the circle x^2+y^2+8x+12y+15 = 0

Answer 1

Answer:

See below.

Step-by-step explanation:

The equation of the circle is,

$x^2 + y^2 + 8x + 12y + 15 = 0$

We rearrange the terms like,

$( x^2 + 8x) + ( y^2 + 12y ) = -15$

Now we complete the squares like,

$(x^2 + 8x + 16 ) + ( y^2 + 12y + 36 ) = -15 + 16 + 36\\( x + 4 )^2  + ( y + 6 )^2 = 37$

We get the standard equation of the circle. The coordinates of the centre of the circle are ( -4 , -6 ).

The circle which passes through ( 2 , 3 ) with centre ( -4 , -6 ) will have a radius of,

$( 2 + 4 )^2 + ( 3 + 6 )^2 = r^2\\r^2 = 117\\r = \sqrt{117}$

We know root 13 is smaller than root 117. Hence, the circle will be concentric if it has a radius of root 117 and centre as ( -4 , -6 ). So the equation of such a circle is,

$( x +4 )^2 + ( y + 6 )^2 = 117$

What we did:

1. Transform the general equation of the circle to its standard form.
2. Deduce the centre coordinates and radius.

Answer 2

Step-by-step explanation:

The equation of the circle is,

x^2 + y^2 + 8x + 12y + 15 = 0x

2

+y

2

+8x+12y+15=0

We rearrange the terms like,

( x^2 + 8x) + ( y^2 + 12y ) = -15(x

2

+8x)+(y

2

+12y)=−15

Now we complete the squares like,

\begin{gathered}(x^2 + 8x + 16 ) + ( y^2 + 12y + 36 ) = -15 + 16 + 36\\( x + 4 )^2 + ( y + 6 )^2 = 37\end{gathered}

(x

2

+8x+16)+(y

2

+12y+36)=−15+16+36

(x+4)

2

+(y+6)

2

=37

We get the standard equation of the circle. The coordinates of the centre of the circle are ( -4 , -6 ).

The circle which passes through ( 2 , 3 ) with centre ( -4 , -6 ) will have a radius of,

\begin{gathered}( 2 + 4 )^2 + ( 3 + 6 )^2 = r^2\\r^2 = 117\\r = \sqrt{117}\end{gathered}

(2+4)

2

+(3+6)

2

=r

2

r

2

=117

r=

117

We know root 13 is smaller than root 117. Hence, the circle will be concentric if it has a radius of root 117 and centre as ( -4 , -6 ). So the equation of such a circle is,

( x +4 )^2 + ( y + 6 )^2 = 117(x+4)

2

+(y+6)

2

=117