Question

The equation of a circle which passes through the three points $(3,0),(1,-6),(4,-1)$ is-

• A. $2x^2+2y^2+5x-11y+3=0$
• B. $x^2+y^{-}5x+11y-3=0$
• C. $x^2+y^2+5x-11y+3=0$
• D. $2x^2+2y^2-5x+11y-3=0$

Answer 1

The correct option is D $2x^2+2y^2-5x+11y-3=0$

Gen equation of a circle is $(x-h{)}^2+(y-k{)}^2=r^2$

$(3,0),(1,-6),(4,-1)$ circle passes through these points

$\therefore$ These three points should satisfy the equation

of circle.

$(3-h{)}^2+k^2=r^2$ ___ (I)

$(1-h{)}^2+(-6-k{)}^2=r^2$

$(1-h{)}^2+(6+k{)}^2=r^2$ ___ (II)

From (I) & (II)

$(3-h{)}^2+k^2=(1-h{)}^2+(6+k{)}^2$

$9+h^2-6h+k^2=1+h^{2}2h+36+k^2+12k$

$9-6h=1-2h+36+12k$

$9-37=4h+12k$

$-28=4h+12k$

$h+3k=-7$

$h=-7-3k$ ___ (III)

$(4-h{)}^2+(-1-k{)}^2=r^2$ ___ (IV)

From (I) & (IV)

$(3-h{)}^2+k^2=(4-h{)}^2+(1+k{)}^2$

$9+h^2-6h+k^2=16+h^2-8h+1+k^2+2k$

$9-6h=17-8h+2k$

$9-17=-2h+2k$

$-8=-2h+2k$

$-h+k=-4$

$k=-4+h$ ___ (V)

Put (V) in (III)

$h=-7-3(-4+h)$

$h=-7+12-3h$

$h=5/4,k=-11/4,r=170/16$

Eq of circle: $(x-5/4{)}^2+(y+11/4{)}^2=170/16$

$x^2+25/16-10x/4+y^2+121/16+22y/4=170/16$

Simplifying, we get

$2x^2+x{y}^2-5x+11y-3=0$