# Conic Sections - Online Test

Q1. If the line 2x – y +  = 0 is a diameter of the circle  then  =
Explaination / Solution:

Equation of circle is

Applying completing the square method

Comparing the above equation with   we get center as (-3,3) and radius as    .

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

=>

=>

Q2. Length of common chord of the circles and  is
Explaination / Solution: Applying the completing the square Comparing the above equation with we get its center as (-1,-3) and radius (r1)= Comparing the above equation with we get its center as (2,1) and radius (r2)=

So distance(d) between the centers is units
After applying general formula of length of common chord as and putting the values of r1, r2 and d we get length as
.

Q3. The circles  and
Explaination / Solution:

Circle touches externally if distance between the centers is equal to the sum of the radii . After applying completing the square, we get so the center is (-3,-3) and radius is After applying completing the square so the center is (6,6) and radius is
Distance between centres and sum of radii are equal.

Hence the circle touches externally.

Q4.

The equation  represents

Explaination / Solution:

The above circle can be written as Here the center is (0,0) and radius is also 0 units.

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

Q5.

The equation represents

Explaination / Solution:

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.

Q6. Circumcentre of the triangle, whose vertices are (0, 0), (6, 0) and (0, 4) is
Explaination / Solution:

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

so mid point of BC=(,) i.e.(3,2)

Q7. If (x-a)+ (y-b)= c2 represents a circle, then
Explaination / Solution:

(x-a)+ (y-b)= 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 c0.

Q8. Two perpendicular tangents to the circle  meet at P. The locus of P is
Explaination / Solution:

locus of P is a circle with centre at origin and radius .This is known as the director circle of the circle
Q9. The number of tangents to the circle which pass through the point ( 3, - 2), is
Explaination / Solution: After completing the square,we get so center is (4,3) and radius is 4.

Distance between center and given point which is greater than 4.

hence point lies outside the circle .

Since point lies outside of the circle there will be 2 tangents since two tangents can be drawn from external point to a circle.

Q10. The length of the chord joining the point ( 4 cos , 4 sin ) and 4 ( cos(+), 4 sin( + )) of the circle  is
Explaination / Solution:

using distance formulae  on simplifying we get, 