Circles Test 22

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Circles Test 22

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the circle passing through the points $$(4, 1), (6, 5)$$ and having the centre on the line $$4x+y-16=0$$ is

A
$${ x }^{ 2 }+{ y }^{ 2 }-6x-8y+15=0$$

B
$$15({ x }^{ 2 }+{ y }^{ 2 })-94x+18y+55=0$$

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

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

Solution

Given points,

$$(4,1),(6,5)$$

equation of circle $$(x-h)^2+(y-k)^2=r^2$$

$$\Rightarrow (4-h)^2+(1-k)^2=r^2$$....(1)

$$\Rightarrow (6-h)^2+(5-k)^2=r^2$$....(2)

solving the above 2 equations, we get,

$$h+2k=11$$....(3)

given, $$4h+k=16$$.....(4)

solving the above 2 equations, we get,

$$h=3,k=4$$

substituting the above values in (1), we get,

$$(4-3)^2+(1-4)^2=r^2$$

$$\therefore r=\sqrt{10}$$

Hence, the equation is,

$$(x-3)^2+(y-4)^2=(\sqrt {10})^2$$

$$x^2+y^2-6x-8y+15=0$$
Question 2
1 / -0

Equation of the circle which is such that the lengths of the tangents to it from the points $$( 1,0 ) , ( 0,2 )$$ and $$( 3,2 )$$ are $$1 , \sqrt { 7 }$$ and $$\sqrt { 2 }$$ respectively is

A
$$6 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$$

B
$$9 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$$

C
$$3 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$$

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

Solution
Question 3
1 / -0

The radius of circle $$x^2+y^2-6x-8y=0$$

A
5

B
4

C
3

D
2

Solution

The radius of circel $$x^2+y^2-6x-8y=0$$ is $$\sqrt{g^2+f^2-c}$$
Here $$g=-3,f=-4,c=0\implies r=\sqrt{(-3)^2+(-4)^2}\\\sqrt{9+16}=\sqrt {25}=5$$
Question 4
1 / -0

If the circle $$x^{2}+y^{2}=9$$ passesthrough $$(2,c)$$ then $$c$$ is equal to

A
$$\sqrt5$$

B
$$\sqrt 6$$

C
$$\sqrt 3$$

D
$$\sqrt 7$$

Solution

The equation of circle $$x^2+y^2=9$$
The point is $$(2,c)$$
$$\implies 2^2+c^2=9\\4+c^2=9\\c^2=9-4\\c^2=5\\c=\sqrt 5$$
Question 5
1 / -0

The equation to the circle which touches the axis of y at the origin and passes through $$(3, 4)$$ is?

A
$$2(x^2+y^2)-3x=0$$

B
$$3(x^2+y^2)-25x=0$$

C
$$4(x^2+y^2)-25y=0$$

D
$$4(x^2+y^2)-25x+10=0$$

Solution
Question 6
1 / -0

The image of the circle $$(x - 3)^2 + (y - 2)^2 = 1$$ in the line mirror $$ax + by = 19$$ is $$(x - 1)^2 + (y - 16)^2 = 1$$ then values of (a, b) is

A
(1,1)

B
(1,1)

C
(1,1)

D
(-1,-1)

Solution
Question 7
1 / -0

The equation of the circle which touches the axis of y at a distance $$+4$$ from the origin and cuts off an intercept $$6$$ from the $$+ve$$ direction of x-axis is

A
$$x^{2}+y^{2}-10x\pm8y-16=0$$

B
$$x^{2}+y^{2}+10x\pm8y+16=0$$

C
$$x^{2}+y^{2}-10x\pm8y+16=0$$

D
$$x^{2}+y^{2}-8x\pm10y-16=0$$

Solution

For a circle with center (a,b) and radius r,
Equation is $$(x-a)^2 + (y-b)^2 = r^2$$

There can be two scenarios for the given conditions as depicted in above figures.

For (A),
In $$\triangle OMN$$, using pythagoras theorem,
$$ON = \sqrt{OM^2 + MN^2} = \sqrt{4^2 + 3^2} = 5$$
So, radius of circle = 5.
And, center = (5,4)

Thus, Equation of cricle
$$\rightarrow$$ $$(x-5)^2 + (y-4)^2 = 5^2$$
$$\rightarrow$$ $$x^2 + y^2 -10x -8y +16 = 0$$

Similarly, For (B).
Equation of circle,
$$\rightarrow$$ $$(x-5)^2 + (y+4)^2 = 5^2$$
$$\rightarrow$$ $$x^2 + y^2 -10x +8y +16 = 0$$
Question 8
1 / -0

The equation of the circle passing through the origin and making intercept $$4,5$$ on the positive coordinates axes is

A
$$x^{2}+y^{2}-4x+5y=0$$

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

C
$$x^{2}+y^{2}+4x+5y=0$$

D
`$$x^{2}+y^{2}+4x-5y=0$$

Solution

For a circle with center (a,b) and radius r,
Equation is $$(x-a)^2 + (y-b)^2 = r^2$$

Their can be two scenarios for given conditions as depicted in above figures.

For (A),
Center = $$(2,\dfrac{5}{2})$$
Radius = $$\sqrt{2^2+(\dfrac{5}{2}})^2 = \dfrac{\sqrt{41}}{2}$$
So, Equation of circle,
$$\rightarrow$$ $$(x-2)^2 + (y-\dfrac{5}{2})^2 = \dfrac{41}{4}$$
$$\rightarrow$$ $$x^2 + y^2 - 4x - 5y = 0$$

Similarly, For (B),
Equation of circle,
$$\rightarrow$$ $$(x-\dfrac{5}{2})^2 + (y-2)^2 = \dfrac{41}{4}$$
$$\rightarrow$$ $$x^2 + y^2 - 4y - 5x = 0$$

Hence, (B) is the correct answer.
Question 9
1 / -0

The equation of circle center at $$(0,0)$$ and Radius $$8cm$$

A
$$x^2+y^2=64cm$$

B
$$x^2+y^2=8$$

C
$$x^2+y^2=16$$

D
$$x^2+y^2=4$$

Solution

The equation of circle is $$x^2+y^2=r^2\\x^2+y^2=8^2\\x^2+y^2=64$$
Question 10
1 / -0

A circle of radius $$'5'$$ touches the coordinate axes in the first quadrant. If the circle makes one complete roll on x-axis along the positive direction, then its equation in new position is

A
$${x}^{2}+{y}^{2}-10(2\pi+1)x-10y+100{\pi}^{2}+100\pi+25=0$$

B
$${x}^{2}+{y}^{2}+10(2\pi+1)x-10y+100{\pi}^{2}+100\pi+25=0$$

C
$${x}^{2}+{y}^{2}-10(2\pi+1)x+10y+100{\pi}^{2}+100\pi+25=0$$

D
$${x}^{2}+{y}^{2}+10(2\pi+1)x+10y+100{\pi}^{2}+100\pi+25=0$$

Solution