Circles Test 24

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Circles Test 24

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Weekly Quiz Competition

Question 1
1 / -0

The value of k, such that the equation
$$2x^{2}+2y^{2}-6x+8y+k=0$$ represent a point circle , is

A
0

B
25

C
$$\frac {25}{2}$$

D
$$-\frac {25}{2}$$

Solution

The given equation of circle is
$$\Rightarrow$$ $$2x^2+2y^2-6x+8y+k=0$$
$$\Rightarrow$$ $$x^2+y^2-3x+4y+\dfrac{k}{2}=0$$ [ Dividing both sides by $$2$$ ]
Comparing it with general equation of circle $$x^2+y^2+2gx+2fy+c=0,$$ we get
$$\Rightarrow$$ $$2g=-3,\,2f=4$$ and $$c=\dfrac{k}{2}$$
$$\therefore$$ $$g=\dfrac{-3}{2}$$ and $$f=2$$
Center $$=(-g,-f)=\left(\dfrac{3}{2},\,-2\right)$$
We know,
Radius $$(r)=\sqrt{g^2+f^2-c}$$
We know that for point circle radius is $$0.$$ $$r=0$$
$$\Rightarrow$$ $$0=\sqrt{\left(\dfrac{3}{2}\right)^2+(-2)^2-\dfrac{k}{2}}$$
$$\Rightarrow$$ $$0=\sqrt{\dfrac{9}{4}+4-\dfrac{k}{2}}$$
$$\Rightarrow$$ $$0=\dfrac{9}{4}+4-\dfrac{k}{2}$$
$$\Rightarrow$$ $$\dfrac{k}{2}=\dfrac{9+16}{4}$$
$$\Rightarrow$$ $$\dfrac{k}{2}=\dfrac{25}{4}$$
$$\therefore$$ $$k=\dfrac{25}{2}$$
Question 2
1 / -0

If the radius of the circle $$x^{2}+y^{2}-18x-12y+k=0$$ be $$11$$ then k=

A
34

B
4

C
-4

D
49

Solution

The equation of circle is $$x^2+y^2-18x-12y+k=0$$.
Comparing it with $$x^2+y^2+2gx+2fy+c=0$$ we get,
$$2g=-18,\,2f=-12$$ and $$c=k$$
$$\therefore$$ $$g=-9,\,f=-6$$ and $$c=k$$
Center $$=(-g,-f)=(9,6)$$
Radius of a circle $$(r)=11$$ [ Given ]
We know,
$$\Rightarrow$$ $$r=\sqrt{g^2+f^2-c}$$
$$\Rightarrow$$ $$11=\sqrt{(-9)^2+(-6)^2-k}$$
$$\Rightarrow$$ $$11=\sqrt{81+36-k}$$
$$\Rightarrow$$ $$11=\sqrt{117-k}$$
$$\Rightarrow$$ $$121=117-k$$
$$\Rightarrow$$ $$4=-k$$
$$\therefore$$ $$k=-4$$
Question 3
1 / -0

If $$P\left(2, 8\right)$$ is an interior point of a circle $${x}^{2}+{y}^{2}–2x + 4y – p = 0$$ which neither touches nor intersects the axes, then set for $$p$$ is ______________.

A
$$p < – 1$$

B
$$p < – 2$$

C
$$p > 2$$

D
$$\phi$$

Solution

For internal point $$P\left(2,8\right)$$
$$4+64-8+32-p<0$$
$$\Rightarrow \,p>92$$ and $$x-$$ intercept$$=2\sqrt{1+p}$$
$$\therefore\,1+p<0$$
$$\Rightarrow \,p<-1$$ and $$y-$$ intercept=2\sqrt{2+p}$$
$$\Rightarrow\,p<-2$$
Question 4
1 / -0

Find the equation of a circle whose centre is (2, - 1 ) an radius is 3

A
$$

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

$$

B
$$

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

$$

C
$$

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

$$

D
$$

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

$$

Solution

Center $$=(2,-1)$$ and $$r=3$$ [ Given ]
We know that, center $$=(-g,-f)$$
So, by comparing we get, $$g=-2$$ and $$f=1$$
$$r=\sqrt{g^2+f^2-c}$$
$$\Rightarrow$$ $$3=\sqrt{(-2)^2+(1)^2-c}$$
$$\Rightarrow$$ $$9=4+1-c$$
$$\Rightarrow$$ $$9=5-c$$
$$\therefore$$ $$c=-4$$
The equation of circle is
$$\Rightarrow$$ $$x^2+y^2+2gx+2fy+c=0$$
$$\Rightarrow$$ $$x^2+y^2+2(-2)x+2(1)y-4=0$$ [ Substituting values of $$g,f$$ and $$c$$ ]
$$\Rightarrow$$ $$x^2+y^2-4x+2y-4=0$$
Question 5
1 / -0

The equation of a circle which is passing through the vertices of an equilateral triangle whose median is of length 3 a is

A
$$

x ^ { 2 } + y ^ { 2 } = 9 a ^ { 2 }

$$

B
$$

x ^ { 2 } + y ^ { 2 } = 16 a ^ { 2 }

$$

C
$$

x ^ { 2 } + y ^ { 2 } = 4 a ^ { 2 }

$$

D
$$

x ^ { 2 } + y ^ { 2 } = a ^ { 2 }

$$

Solution

Median of the equilateral triangle is $$3a$$
$$\Rightarrow=3a$$
In $$\triangle ADB$$,we have
$$\Rightarrow(AB)^2=(AD)^2+(BD)^2$$
$$\Rightarrow(AB)^2=(3a)^2+\left(\dfrac{AB}{2}\right)^2$$
$$\Rightarrow(AB)^2=9a^2+\dfrac{(AB)^2}{4}$$
$$\Rightarrow\dfrac{3}{4}(AB)^2=9a^2$$
$$\Rightarrow(AB)^2=12a^2$$
Now, In $$\triangle OBD,$$
$$\Rightarrow(OB)^2=(OD)^2+(BD)^2$$
$$\Rightarrow r^2=(3a-r)^2+\left(\dfrac{AB}{2}\right)^2$$
$$\Rightarrow r^2=9a^2+r^2-6ar+3a^2$$ $$[\because(AB)^2=10a^2]$$
$$\Rightarrow6ar=12a^2$$
$$\Rightarrow r=2a$$
So, the equation of circle is
$$\Rightarrow(x-0)^2+(y-0)^2=(2a^2)$$
$$\Rightarrow x^2+y^2=4a^2$$
Question 6
1 / -0

A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end of the diameter through A is?

A
$$(x-p)^2=4qy$$

B
$$(x-q)^2=4py$$

C
$$(y-p)^2=4qx$$

D
$$(y-q)^2=4px$$

Solution

$$\Rightarrow (h-p)^2+(k-q)=4\left(\dfrac{k+q}{2}\right)^2$$
$$=k^2+2kq+q^2$$
$$\Rightarrow (x-p)^2=4qy$$.
Question 7
1 / -0

$$AB$$ is a chord of the circle $$x^{2} + y^{2} = 9$$. The tangent at $$A$$ and $$B$$ intersect at $$C$$. If $$(1, 2)$$ is the midpoint of $$AB$$, the area of $$\triangle ABC$$ is (in square units).

A
$$9$$

B
$$\dfrac {7}{\sqrt {5}}$$

C
$$\dfrac {9}{\sqrt {5}}$$

D
$$\dfrac {8}{\sqrt {5}}$$

Solution

The equation of chord with midpoint $$(1, 2)$$ in the circle $$x^{2} + y^{2} = 9$$ is
$$x \times 1 + y\times 2 - 9 = 1^{2} + 2^{2} - 9$$
$$\Rightarrow x + 2y - 5 = 0 .... (1)$$
Also, if the point of intersection of tangents $$C$$ has coordinates $$(h, k)$$, then the equation of common chord is
$$hx + ky - 9 = 0 .... (2)$$
Since (1) and (2) represent the same equation, we get
$$\dfrac {h}{1} = \dfrac {k}{2} = \dfrac {9}{5}$$
$$\Rightarrow h = \dfrac {9}{5}, k = \dfrac {18}{5}$$
From figure, $$AD = \sqrt {OA^{2} - OD^{2}} = \sqrt {9 - 5} = 2$$
$$CD = OC - OD = \sqrt {h^{2} + k^{2}} - \sqrt {5} = \dfrac {9}{\sqrt {5}} - \sqrt {5} = \dfrac {4}{\sqrt {5}}$$
The area of $$ABC = \dfrac {1}{2} AB . CD = AD.CD = 2\times \dfrac {4}{\sqrt {5}} = \dfrac {8}{\sqrt {5}}$$.
Question 8
1 / -0

A circle whose centre is the point of intersection of the lines $$2x-3y+4=0$$ and $$3x+4y-5=0$$ passes through the origin. Find its equation.

A
$$\left(x+\dfrac{1}{17}\right)^{2}+\left(y-\dfrac{22}{17}\right)^{2}=\dfrac{485}{289}$$

B
$$\left(x+\dfrac{1}{17}\right)^{2}+\left(y+\dfrac{21}{17}\right)^{2}=\dfrac{458}{289}$$

C
$$\left(x-\dfrac{2}{17}\right)^{2}+\left(y-\dfrac{22}{17}\right)^{2}=\dfrac{485}{297}$$

D
None of these

Solution

Given, $$2x-3y+4=0$$......(i)
$$3x+4y-5=0$$......(ii)

According to question, intersection of eqn (i) and eqn (ii) is the centre of required circle.

$$\therefore$$ solving (i) & (ii), we get $$x=\dfrac{-1}{17}, y=\dfrac{22}{17}$$
Distance from centre to origin, will give us the radius.

$$\therefore D=\sqrt{\left(\dfrac{-1}{17}-0\right)^2+\left(\dfrac{22}{17}-0\right)^2}$$

$$r=\sqrt{\dfrac{1}{289}+\dfrac{484}{289}}$$
$$=\sqrt{\dfrac{485}{289}}$$

$$\therefore r=\dfrac{\sqrt{485}}{17}$$

Now, eqn of circle $$(x-h)^2+(y-k)^2=r^2$$

$$\Rightarrow \left(x+\dfrac{1}{17}\right)^2+\left(y-\dfrac{22}{17}\right)^2=\dfrac{485}{289}$$
Question 9
1 / -0

The equation of the circle with centre $$(2, 2)$$ which passes through $$(4,5)$$ is

A
$$x^2 + y^2 - 4x + 4y - 77 = 0$$

B
$$x^2 + y^2 - 4x - 4y - 5 = 0$$

C
$$x^2 + y^2 + 2x + 2y - 59 = 0$$

D
$$x^2 + y^2 - 2x - 2y - 23 = 0$$

E
$$x^2 + y^2 + 4x - 2y -26 = 0$$

Solution

Radius of circle is $$\sqrt{(4-2)^2+(5-2)^2}=\sqrt{13}$$ So, equation of circle is
$$(x-2)^2+(y-2)^2=13$$
$$\Rightarrow x^2+4-4x+y^2+4-4y=13$$
$$\Rightarrow x^2+y^2-4x-4y-5=0$$
Question 10
1 / -0

The centre of a circle is $$C(2,-5)$$ and the circle passes through the point $$A(3,2)$$. The equation of the circle is

A
$${ x }^{ 2 }+{ y }^{ 2 }-4x+10y-21=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }+4x+6y-21=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }+4x-10y+21=0$$

D
none of these

Solution

radius of the circle is $$AC=\sqrt { { (3-2) }^{ 2 }+{ (2+5) }^{ 2 } } =\sqrt { 1+49 } =\sqrt { 50 } $$
equation of the circle is
$${ (x-x_1) }^{ 2 }+{ (y-y_1) }^{ 2 }=r^2$$
$${ (x-2) }^{ 2 }+{ (y+5) }^{ 2 }=50\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x+10y-21=0$$

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Re-Attempt Weekly Quiz Competition

Circles Test 24

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Circles Test 24

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SHARING IS CARING

Question 1

0

25

$$\frac {25}{2}$$

$$-\frac {25}{2}$$

Solution

Question 2

34

4

-4

49

Solution

Question 3

$$p < – 1$$

$$p < – 2$$

$$p > 2$$

$$\phi$$

Solution

Question 4

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

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

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

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

Solution

Question 5

$$x ^ { 2 } + y ^ { 2 } = 9 a ^ { 2 }$$

$$x ^ { 2 } + y ^ { 2 } = 16 a ^ { 2 }$$

$$x ^ { 2 } + y ^ { 2 } = 4 a ^ { 2 }$$

$$x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$$

Solution

Question 6

$$(x-p)^2=4qy$$

$$(x-q)^2=4py$$

$$(y-p)^2=4qx$$

$$(y-q)^2=4px$$

Solution

Question 7

$$9$$

$$\dfrac {7}{\sqrt {5}}$$

$$\dfrac {9}{\sqrt {5}}$$

$$\dfrac {8}{\sqrt {5}}$$

Solution

Question 8

$$\left(x+\dfrac{1}{17}\right)^{2}+\left(y-\dfrac{22}{17}\right)^{2}=\dfrac{485}{289}$$

$$\left(x+\dfrac{1}{17}\right)^{2}+\left(y+\dfrac{21}{17}\right)^{2}=\dfrac{458}{289}$$

$$\left(x-\dfrac{2}{17}\right)^{2}+\left(y-\dfrac{22}{17}\right)^{2}=\dfrac{485}{297}$$

None of these

Solution

Question 9

$$x^2 + y^2 - 4x + 4y - 77 = 0$$

$$x^2 + y^2 - 4x - 4y - 5 = 0$$

$$x^2 + y^2 + 2x + 2y - 59 = 0$$

$$x^2 + y^2 - 2x - 2y - 23 = 0$$

$$x^2 + y^2 + 4x - 2y -26 = 0$$

Solution

Question 10

$${ x }^{ 2 }+{ y }^{ 2 }-4x+10y-21=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+4x+6y-21=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+4x-10y+21=0$$

none of these

Solution

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$$

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

$$

$$

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

$$

$$

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

$$

$$

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

$$

$$

x ^ { 2 } + y ^ { 2 } = 9 a ^ { 2 }

$$

$$

x ^ { 2 } + y ^ { 2 } = 16 a ^ { 2 }

$$

$$

x ^ { 2 } + y ^ { 2 } = 4 a ^ { 2 }

$$

$$

x ^ { 2 } + y ^ { 2 } = a ^ { 2 }

$$