Circles Test 19

Result

Circles Test 19

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Weekly Quiz Competition

Question 1
1 / -0

If a circle with centre $$(0,0)$$ touches the line $$5x+12y=1$$ then it equation will be

A
$$169(x^{2}+y^{2})=1$$

B
$$ (x^{2}+y^{2})=169$$

C
$$16(x^{2}+y^{2})=1$$

D
$$ (x^{2}+y^{2})=13$$

Solution
Question 2
1 / -0

The equation of the circle in the first quadrature touching each coordinate axis at a distance of one unit from the origin

A
$$x^{2}+y^{2}-2x-2y+1=0$$

B
$$x^{2}+y^{2}-2x-1=0$$

C
$$x^{2}+y^{2}-2x=0$$

D
None of these

Solution

We can clearly see from diagram that co-ordinates of centre of circle $$=(1,1)$$ & radius is unit
Thus equation of circle:-
$$(x-1)^{2}+(y-1)^{2}=1^{2}$$
$$x^{2}+y^{2}-2x-2y+1=0$$
Question 3
1 / -0

A circle has its centre on the $$y-axis$$ and passes through the origin, touches another circle with centre $$(2,2)$$ and radius 2, then the radius of the circle is

A
$$1$$

B
$$1/2$$

C
$$1/3$$

D
$$1/4$$

Solution
Question 4
1 / -0

The equation of the circle which circumscribes the triangle formed by the lines x + y + 3 = 0, x - y + 1 =0 and x = 3 is

A
$$x^2+y^2-6x+2y-15=0$$

B
$$3x^2+3y^2-9=0$$

C
$$x^2+y^2+6x-2y+15=0$$

D
None of these

Solution
Question 5
1 / -0

The equation of the circle which touches x-axis and whose center is (1,2) is

A
$$x^{2}+y^{2}-2x+4y+1=0$$

B
$$x^{2}+y^{2}-2x-4y+1=0$$

C
$$x^{2}+y^{2}+2x+4y+1=0$$

D
$$x^{2}+y^{2}-4x+2y+4=0$$

Solution

$$\therefore O\equiv (1,2) \ \ r=2cm$$
$$h,k$$
$$\therefore$$ Eqn of circle:
$$(x-h)^2+(y-k)^2=r^2$$
$$(x-1)^2+(y-2)^2=4$$
$$x^2+1-2x+y^2+4-4y=4$$
$$x^2+y^2-2x-4y+1=0$$
Question 6
1 / -0

The parametric form of the equation of the circle $${ x }^{ 2 }+{ y }^{ 2 }=9$$ is:

A
$$x=\sqrt { 3 } \cos { \theta } ,$$ $$y=\sqrt { 3 } \sin { \theta } $$

B
$$x=3\cos { \theta } ,\quad y=3\sin { \theta } $$

C
$$x=-\sqrt { 3 } \cos { \theta } ,$$ $$y=-\sqrt { 3 } \sin { \theta } $$

D
none of these

Solution
Question 7
1 / -0

The equation of a circle with centre at $$(1, -2)$$ and passing through the centre of the given circle $$x^2+y^2+2y-3=0$$, is?

A
$$x^2+y^2-2x+4y+3=0$$

B
$$x^2+y^2-2x+4y-3=0$$

C
$$x^2+y^2+2x-4y-3=0$$

D
$$x^2+y^2+2x-4y+3=0$$

Solution
Question 8
1 / -0

The equation of circle whose centre is $$(1, -3)$$ and which touches the line $$2x-y-4=0$$, is

A
$$5x^2+5y^2+10x+30y+49=0$$

B
$$5x^2+5y^2+10x-30y-49=0$$

C
$$5x^2+5y^2-10x+30y-49=0$$

D
None of these

Solution
Question 9
1 / -0

$$ABCD$$ is a square with side $$a$$. If $$AB$$ and $$AD$$ are taken as positive coordinate axes then equation of circle circumscribing the square is

A
$$x^{2}+y^{2}-ax-ay=0$$

B
$$x^{2}+y^{2}+ax+ay=0$$

C
$$x^{2}+y^{2}-ax+ay=0$$

D
$$x^{2}+y^{2}+ax-ay=0$$

Solution

Let the radius of circle be a/2, since it is
half of the diameter which is a. The centre of
the circle is going to occur at (a/2, a/2), since that
is also the centre of square
Using $$ (x-h)^{2}+(y-k)^{2} = r^{2}$$
where r is the radius and (h,k) is centre
$$ (x-a/2)^{2}+(y-a/2)^{2} = (a/2)^{2}$$
$$ x^{2}-\frac{2ax}{2}+\frac{a^{2}}{4}+y^{2}-\frac{2ay}{2}+\frac{a^{2}}{4} = \frac{a^{2}}{2}$$
$$ x^{2}+y^{2}+\frac{a^{2}}{2}-ax - ay = \frac{a^{2}}{2}$$
$$ x^{2}+y^{2} = ax+ay.$$
$$ x^{2}+y^{2} = a(x+y) \Rightarrow x^{2}+y^{2}-ax-ay = 0.$$
Question 10
1 / -0

The circle passing through $$\left(t,1\right),\left(1,t\right)$$ and $$\left(t,t\right)$$ for all values of $$t$$ also passes through

A
$$\left(0,0\right)$$

B
$$\left(1,1\right)$$

C
$$\left(1,-1\right)$$

D
$$\left(-1,-1\right)$$

Solution

The general equation of circle is
$$(x-h)^{2}+(y-k)^{2} = r^{2},$$ where h and k are the
coordinates of center of circle.
if It passes through $$(1,t)$$ so $$ (1-h)^{2}+(t-k)^{2} = r^{2}$$ __ (1)
if It passes through $$(t,t)$$ so $$ (t-h)^{2}+(t-k)^{2} = r^{2}$$ __ (2)
if It passes through $$(t,1)$$ so $$ (t-h)^{2}+(1-k)^{2} = r^{2}$$ __ (3)
on solving eq (1) & (2)
$$ 1-h = t-h \Rightarrow t = 1 $$
It t = 1, then the circle passes through (1,1).