Question

Column 1 has the condition on circle and column 2 has the equation of circles.which one of them are matched correct?

Column 1 column 2

1) Centre of origin and radius = 10 P) $(x-8)(x+8)+(y-6)(y+6)=0$

2) Extremeties of diameter are (8,6) and (-8,-6) Q) $(x-6)(x+6)+(y-8)(y+8)=0$

3) Centre is (1,-2) and radius is 4 R) $x^2+y^2-2x+4y-11=0$

Answer 1

1) Equation of circle with origin as centre and radius r is given by $x^2+y^2=r^2$
$\Rightarrow x^2+y^2=100$ is the equation of the circle we will expand the equation given P and check if it gives same equation $x^2+y^2=100$ .
$(x-8)(x+8)+(y-6)y+6=0$
$\Rightarrow x^2-64+y^2-36=0$
$\Rightarrow x^2+y^2=100$
So both of them give the same equation

2) Equation of circle with $(x_1,y_1)$ and $(x_2,y_2)$ as the extremeties of diameter is given by
$(x-x_1)(x,x_2)+(y-y_1)(y-y_2)$
$\Rightarrow (x-8)(x+8)+(y-6)(y+6)=0$
$\Rightarrow (x^2+y^2=100$
We will expand Q and see it gives the same equation.
$(x-6)(x+6)+(y-8)(y+8)=0$
$\Rightarrow x^2+y^2=100$
$\Rightarrow$ 2 is correctly matched

3) Equation of a circle with centre $(h,k)$ and radius r is given by
$(x-h{)}^2+(y-k{)}^2=r^2$
$\Rightarrow (x-1{)}^2+(y+2{)}^2=4^2$
$\Rightarrow x^2+y^2-2x+4y+1+4=16$
$\Rightarrow x^2+y^2-2x+4y-11=0$

$\Rightarrow$ 3 is correctly matched.