Conic Sections - Online Test

Equation of circle is x2+y2+6x−6y+5=0

x2+6x+y2−6y=−5

Applying completing the square method

(x+3)2+(y−3)2=(13−−√)2

Comparing the above equation with(x−h)2+(y−k)2=(r)2   we get center as (-3,3) and radius as 13−−√   .

As centre of the circlre lies on diameter , it will satisfy the equation of diameter, so on putting (-3,3) in equation of diameter we get

2(−3)−(3)+λ=0

=> −6−3−λ=0

=>λ=9

Distance between centres  and sum of radii   are equal.

Hence the circle touches externally.

So it is a degenerate circle as  degenerate circle is a circle( a point) where radius is zero units.

The general equation of the circle is x2+y2-2gh-2fy+c = 0. Sice the given equations satisfies the general equation, it represents the equation of the circle.

circumcentre of a right angled triangle ABC right angled at A is b+c2 as circumcentre of right angled triangle lies on the mid pont of the hypotenuse.

so mid point of BC=(6+02,0+42) i.e.(3,2)

(x-a)2 + (y-b)2 = c2  here (a,b) is center and c is the radius

and radius cannot be zero because if radius is zero it will become a point or degenerate circle so c≠0.