Circles Test 28

Result

Circles Test 28

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Weekly Quiz Competition

Question 1
1 / -0

Equation of the circle whose radius is 3 and centre is $$(-1,2)$$

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

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

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

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

Solution
Question 2
1 / -0

The equation of circle at center $$(3,4)$$ and radius $$5$$.

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

B
$$x^2+y^2+3x+4y+5=0$$

C
$$x^2+y^2+6x+8y=0$$

D
None.

Solution

The center of equation $$(3,4)$$
The radius of circle is $$5$$
The equation of circle is
$$(x-3)^2+(y-4)^2=5^2\\x^2+y^2-6x+9-8y+16=25\\x^2+y^2-6x-8y=0$$
Question 3
1 / -0

The product of perpendicular drawn from the origin to the lines represented by the equation $$ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$$, will be:

A
$$\dfrac{ab}{\sqrt{a^{2}-b^{2}+4h^{2}}}$$

B
$$\dfrac{bc}{\sqrt{a^{2}-b^{2}+4h^{2}}}$$

C
$$\dfrac{ca}{\sqrt{a^{2}-b^{2}+4h^{2}}}$$

D
$$\dfrac{c}{\sqrt{(a-b)^{2}+4h^{2}}}$$

Solution

Let $${ l }_{ 1 }x+{ m }_{ 1 }y+{ n }_{ 1 }=0$$ & $${ l }_{ 2 }x+{ m }_{ 2 }y+{ n }_{ 2 }=\theta $$ be two lines represented by $${ ax }^{ 2 }+2hxy+{ by }^{ 2 }+2gx+2fy+c=0$$
$${ l }_{ 1 }{ l }_{ 2 }=a;\left( { l }_{ 1 }{ m }_{ 2 }+{ l }_{ 2 }{ n }_{ 1 } \right) =2h$$ , $${ m }_{ 1 }{ m }_{ 2 }=b,\left( { l }_{ 1 }{ n }_{ 2 }+{ l }_{ 2 }{ n }_{ 1 } \right) =2g,\left( { m }_{ 1 }{ n }_{ 2 }+{ m }_{ 2 }{ n }_{ 1 } \right) =2f$$
$${ n }_{ 1 }{ n }_{ 2 }=c$$
Perpendicular from origin to $${ l }_{ 1 }x+{ m }_{ 1 }y+{ n }_{ 1 }=0$$
$$=\dfrac { { n }_{ 1 } }{ \sqrt { { l }_{ 1 }^{ 2 }+{ m }_{ 1 }^{ 2 } } } $$
Perpendicular from origin to $${ l }_{ 2 }x+{ m }_{ 2 }y+{ n }_{ 2 }=0$$
$$=\dfrac { { n }_{ 2 } }{ \sqrt { { l }_{ 2 }^{ 2 }+{ m }_{ 2 }^{ 2 } } } $$
Product, $$\dfrac { { n }_{ 1 }{ n }_{ 2 } }{ \sqrt { \left( { l }_{ 1 }^{ 2 }+{ m }_{ 1 }^{ 2 } \right) \left( { l }_{ 2 }^{ 2 }+{ m }_{ 2 }^{ 2 } \right) } } $$
$$=\dfrac { C }{ \sqrt { { \left( { l }_{ 1 }{ l }_{ 2 } \right) }^{ 2 }+\left( { l }_{ 1 }^{ 2 }{ m }_{ 2 }^{ 2 }+{ l }_{ 2 }^{ 2 }{ m }_{ 1 }^{ 2 } \right) +{ \left( { m }_{ 1 }{ m }_{ 2 } \right) }^{ 2 } } } $$
$$=\dfrac { C }{ \sqrt { { \left( { l }_{ 1 }{ l }_{ 2 } \right) }^{ 2 }+{ \left( { l }_{ 1 }{ m }_{ 2 }+{ l }_{ 2 }{ m }_{ 1 } \right) }^{ 2 }{ -2{ l }_{ 1 }{ l }_{ 2 }{ m }_{ 1 }{ m }_{ 2 }+\left( { m }_{ 1 }{ m }_{ 2 } \right) }^{ 2 } } } $$
$$=\dfrac { C }{ \sqrt { { a }^{ 2 }+{ 4h }^{ 2 }-2ab+{ b }^{ 2 } } } $$
$$=\dfrac { C }{ \sqrt { { \left( a-b \right) }^{ 2 }+{ 4h }^{ 2 } } } $$
Question 4
1 / -0

The equation of the image of the circle $${ \left( x-3 \right) }^{ 2 }\quad +\quad { \left( y-2 \right) }^{ 2 }\quad =1\quad in\quad the\quad mirror\quad x+y=19\quad is$$

A
$${ \left( x-14 \right) }^{ 2 }\quad +\quad { \left( y-13 \right) }^{ 2 }\quad =1$$

B
$${ \left( x-15 \right) }^{ 2 }\quad +\quad { \left( y-14 \right) }^{ 2 }\quad =1$$

C
$${ \left( x-16 \right) }^{ 2 }\quad +\quad { \left( y-15 \right) }^{ 2 }\quad =1$$

D
$${ \left( x-17 \right) }^{ 2 }+{ \left( y-16 \right) }^{ 2 }\quad =1$$

Solution
Question 5
1 / -0

If the equation $$p{ x }^{ 2 }+(2-q)xy+3{ y }^{ 2 }-6qx+30y+6q=0$$ represents a circle, then the values of p and q are

A
3, 1

B
2, 2

C
3, 2

D
3, 4

Solution
Question 6
1 / -0

Equation of the circle whose radius is $$a+b$$ and centre $$(a, -b)$$

A
$${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=2ab$$

B
$${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=-2ab$$

C
$${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=ab$$

D
$${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=-ab$$

Solution
Question 7
1 / -0

If $$y ^ { 2 } - 2 x - 2 y + 5 = 0$$ is

A
a circle with centre $$( 1,1 )$$

B
a parabola with vertex

C
a parabola with directrix $$x = 3 / 2$$

D
a parabola with directrix
Question 8
1 / -0

A circle passes through $$A(1,2)$$ and the equations of the normal to the circle is $$x+2y=5$$. If the circle passes through $$B(-5,5)$$, then the radius of the circle is

A
$$\dfrac {3}{2}$$

B
$$\dfrac {\sqrt {5}}{2}$$

C
$$\dfrac {3\sqrt {5}}{2}$$

D
$$\dfrac {5}{2}$$

Solution
Question 9
1 / -0

The equation of circle with centre (1, 2) and tangent $$x + y - 5 = 0$$ is

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

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

C
$${x^2} + {y^2} - 2x + 4y + 8 = 0$$

D
$${x^2} + {y^2} - 2x - 4y + 8 = 0$$

Solution
Question 10
1 / -0

If $$ x^{2}+y^{2}-2 x+2 a y+a+3=0 $$ represents a real circle with non-zero radius, then most appropriate is

A
$$

a \in(-\infty,-0)

$$

B
$$

a \in(-1,2)

$$

C
$$

a \in(2,\infty)

$$

D
$$

a \in(-\infty,-1) \cup(2 \infty)

$$

Solution