Circles Test 15

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Circles Test 15

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Weekly Quiz Competition

Question 1
1 / -0

The radius of the circle $$x^2+y^2-5x+2y+5=0$$ is

A
$$2$$

B
$$1$$

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

D
$$\dfrac{2}{3}$$

Solution

Given the equation of the circle is $$x^2+y^2-5x+2y+5=0$$ can be written as
or $$x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}+y^2+2.y.1+1+5-1-\dfrac{25}{4}=0$$
or $$\left(x+\dfrac{5}{2}\right)^2+(y+1)^2+4-\dfrac{25}{4}=0$$
or $$\left(x+\dfrac{5}{2}\right)^2+(y+1)^2=\dfrac{9}{4}$$
or $$\left(x+\dfrac{5}{2}\right)^2+(y+1)^2=\left(\dfrac{3}{2}\right)^2$$.
From this equation it is clear that the centre of the circle is $$\left(-\dfrac{5}{2},-1\right)$$ and radius is $$\dfrac{3}{2}$$.
Question 2
1 / -0

The equation of the circle passing through $$(3,6)$$ and whose centre is $$(2,-1)$$ is

A
$${x^2} + {y^2} - 4x + 3y = 45$$

B
$${x^2} + {y^2} - 4x + 2y - 45 = 0$$

C
$${x^2} + {y^2} + 4x - 2y = 45$$

D
$${x^2} + {y^2} - 4x + 2y + 45 = 0$$

Solution

$$\textbf{Step-1: Finding radius(r)}$$

               $$\text{ We can find radius by calculating distance between centre and circumference point from distance formula}$$

               $$\text{ Distance between two points $$(x1,y1) , (x2,y2)}$$

                $$\text{Is given as D}$$ = $$\sqrt {{{(x_1 - x_2)}^{^2}} + {{(y_1 - y_2)}^2}} $$, Here x1 = 2, x2 = 3, y1 = -1, y2 = 0

                $$\text{So r comes out to}$$
                be r =$$\sqrt{2}$$

$$\textbf{Step-2: Required equation of circle is given by}$$ $${(x - x_1)^2} + {(y - y_!)^2} = {r^2}$$

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

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

$$\textbf{Hence the required equation of crcle is}$$ $${x^2} + {y^2} - 4x + 2y + 3 = 0$$
Question 3
1 / -0

Cantres of the three circles
$$x^{2}+y^{2}-4x-6y-14=0$$
$$x^{2}+y^{2}+2x+4y-5=0$$
and $$x^{2}+y^{2}-10x-16y+7=0$$

A
Are the vertices of a right triangle

B
The vertices of an isosceles triangle which is not regular

C
Vertices of a regular triangle

D
Are collinear

Solution

Centre of $$x^2+y^2-4x-6y-14=0\space is\space (-2,-3)$$
Centre of $$x^2+y^2+2x+4y-5=0\space is\space (1,2)$$
Centre of $$x^2+y^2-10x-16y+7=0\space is\space (-5,-8)$$

Let these centres be $$A(-2,-3);B(1,2);C=(-5,-8)$$

Now we find the lengths of $$AB, BC, CA$$

$$\Rightarrow AB=\sqrt{(-2-1)^2+(-3-2)^2}=\sqrt {34}$$

$$\Rightarrow BC=\sqrt{(-5-1)^2+(-8-2)^2}=\sqrt {136}$$

$$\Rightarrow CA=\sqrt{(-2+5)^2+(-3+8)^2}=\sqrt {34}$$

Here we can see that $$AB+CA=BC$$

Therefore $$ ABC $$ is a straight line.
Question 4
1 / -0

The centre of the circle $$(x-a)(x-b)+(y-c)(y-c)=0$$ is

A
$$(a+b,c+d)$$

B
$$9-a-b,-c-d)$$

C
$$\left(\dfrac{a+b}{2},\dfrac{c+d}{2}\right)$$

D
none of these

Solution

Given the equation of the circle is
$$(x-a)(x-b)+(y-c)(y-c)=0$$
or, $$x^2-x(a+b)+ab+y^2-x(c+c)+cd=0$$
or, $$x^2-2.x.\dfrac{a+b}{2}+\dfrac{(a+b)^2}{4}+y^2-2.y.\dfrac{c+d}{2}+\dfrac{(c+d)^2}{4}=\dfrac{(a+b)^2}{4}-ab+\dfrac{(c+d)^2}{4}-cd$$
or, $$\left(x-\dfrac{a+b}{2}\right)^2+\left(y-\dfrac{c+d}{2}\right)^2=\dfrac{(a-b)^2}{4}+\dfrac{(c-d)^2}{4}$$.
From this equation it is clear that the centre is $$\left(\dfrac{a+b}{2},\dfrac{c+d}{2}\right)$$.
Question 5
1 / -0

If a be the radius of a circle which touches x-axis at the origin, then its equation is

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

B
$${ x }^{ 2 }+{ y }^{ 2 }\pm 2ya=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }\pm 2xa=0$$

D
$${ x }^{ 2 }+{ y }^{ 2 }+ya=0$$

Solution

The equation of the circle with centre at $$\left(h,k\right)$$ and radius equal to $$a$$ is $${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$$
When the circle passes through the origin and centre lies on $$x-$$ axis
$$\Rightarrow h=a$$ and $$k=0$$
Then the equation $${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$$ becomes $${\left(x-a\right)}^{2}+{y}^{2}={a}^{2}$$
If a circle passes through the origin and centre lies on $$x-$$axis then the abscissa will be equal to the radius of the circle and the $$y-$$co-ordinate of the centre will be zero Hence, the equation of the circle will be of the form
$${\left(x\pm a\right)}^{2}+{y}^{2}={a}^{2}$$
$$\Rightarrow {x}^{2}+{a}^{2}\pm 2ax+{y}^{2}={a}^{2}={x}^{2}+{y}^{2}\pm 2ax=0$$ is the required equation of the circle.
Question 6
1 / -0

Equation of the circle with centre on the $$y-$$axis and passing through the origin and the point $$(2,3)$$ is

A
$$x^{2}+y^{2}+13y=0$$

B
$$3x^{2}+3y^{2}+13x+3=0$$

C
$$6x^{2}+6y^{2}-13y=0$$

D
$$x^{2}+y^{2}+13+3=0$$

Solution

We have to find equation of a circle with center on the y-axis.
General equation of such circle is
$$(x-0)^{2}+(y-k)^{2}=k^{2}$$
It passes through $$(2,3)$$
i.e, $$2^{2}+(3-k)^{2}=k^{2}$$
$$\Rightarrow 4+9+k^{2}-6k=k^{2}$$
$$\Rightarrow k=\dfrac{13}{6}$$
$$\therefore $$ Equation of circle $$=x^{2}+\left(y-\dfrac{13}{6}\right)^{2}=\left(\dfrac{13}{6}\right)^{2}$$
$$=6x^{2}+6y^{2}-13y=0$$
Question 7
1 / -0

The graph of curve
$${x^2} + {y^2} - 8x - 8y + 32 = 0$$ falls wholly in the

A
first quadrant

B
second quadrant

C
third quadrant

D
none of these

Solution

$${x}^{2}+{y}^{2}−2xy−8x−8y+32=0$$
$$\Rightarrow\,{x}^{2}+{y}^{2}−2xy=8x+8y-32$$
$$\Rightarrow {\left(x-y\right)}^{2}= 8\left(x + y - 4\right)$$
is a parabola whose axis is $$x - y = 0$$ and the tangent at the vertex is $$x + y - 4 = 0$$
Also, when $$y = 0$$, we have
$$x^2 - 8x + 32 = 0$$ which gives no real values of $$x$$
when$$ x = 0$$, we have $$y^2 - 8y + 32 = 0$$ which gives no real values of $$y$$.
So, the parabola does not intersect the axes. Hence, the graph falls in the first quadrant.
Question 8
1 / -0

$$(a, c)$$ and $$(b, c)$$ are the centres of two circles whose radical axis is the y-axis. If the radius of first circle is $$r$$ then the diameter of the other circle is

A
$$2\sqrt {{r^2} - {b^2} + {a^2}} $$

B
$$\sqrt {{r^2} - {a^2} + {b^2}} $$

C
$$\left( {{r^2} - {b^2} + {a^2}} \right)$$

D
$$2\sqrt {{r^2} - {a^2} + {b^2}} $$

Solution

Let radius of after circle be $$R$$
$$C_1:(x-a)^2+(y-c)^2=r^2$$
$$C_2:(x-b)^2+(y-c)^2=R^2$$
Radius axis $$\to x=0$$ OR $$c_1-c_2=0$$
$$c_1-c_2=x$$
$$2bx-2ax-r^2+R^2+a^2-b^2=x$$
Put $$x=0$$
$$R^2=r^2+b^2-a^2$$
$$R=\sqrt {r^2+b^2-a^2}$$
Diameter $$=2\sqrt {r^2 +b^2 -a^2}$$
$$D$$ is correct
Question 9
1 / -0

If $$(4,3)$$ and $$(-12,-1)$$ are end points of a diameter of a circle, then the equation of the circle is-

A
$${ x }^{ 2 }+{ y }^{ 2 }-8x-2y-51=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }+8x-2y-51=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }+8x+2y-51=0$$

D
None of these

Solution

End points of diameter are $$(4,3)$$ and $$(-12,-1)$$
$$\therefore $$ Center of circle$$=\left( \cfrac { 4-12 }{ 2 } ,\cfrac { 3-1 }{ 2 } \right) $$
$$=(-4,1)$$
Radius$$=\cfrac { 1 }{ 2 } \sqrt { { \left( 4+12 \right) }^{ 2 }+{ \left( 3+1 \right) }^{ 2 } } $$
$$=\sqrt { 68 } $$
$$\therefore $$ Equation of circle is $${ \left( x+4 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 }={ \left( \sqrt { 68 } \right) }^{ 2 }$$
$${ x }^{ 2 }+8x+16+{ y }^{ 2 }-2y+1=68$$
$${ x }^{ 2 }+{ y }^{ 2 }+8x-2y-51=0$$
Question 10
1 / -0

if the lines $$ 3x-4y-7=0$$ and $$2x-3y-5=0$$ are two diameter of a circle of area $$49\pi$$ square units the equation of the circle is

A
$$ x^2+y^2+2x-2y-62=0$$

B
$$ x^2+y^2-2x+2y-62=0$$

C
$$ x^2+y^2-2x+2y-47=0$$

D
$$ x^2+y^2+2x-2y-47=0$$

Solution

Given that,
Equations of diameter $$3x-4y-7=0$$ …… (1) and $$2x-3y-5=0$$…… (2)
Solve equations are
$$3x-4y-7=0$$ $$×2$$
$$2x-3y-5=0$$ $$×3$$
$$ 6x-8y-14=0 $$
$$ \underline{6x-9y-15=0}\,\,\,\,\,\,\,\,on\,\,subtracting\,\,\,that $$
$$ y+1=0 $$
$$ y=-1 $$
Put the value of in equation (1)
$$3x-4y-7=0$$
$$ 3x-4\left( -1 \right)-7=0 $$
$$ 3x+4-7=0 $$
$$ 3x-3=0 $$
$$ x=1 $$
Hence , the coordinates of the centre is $$(1,-1)$$
Also given area of circle= $$49\pi$$
$$ \pi {{r}^{2}}=49\pi $$
$$ {{r}^{2}}=49 $$
$$ r=7 $$
Then, we know that the equation of circle is
$$ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} $$
$$ {{\left( x-1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{7}^{2}} $$
$$ {{x}^{2}}+1-2x+{{y}^{2}}+1+2y=49 $$
$$ {{x}^{2}}+{{y}^{2}}-2x+2y+2=49 $$
$$ {{x}^{2}}+{{y}^{2}}-2x+2y=49-2 $$
$$ {{x}^{2}}+{{y}^{2}}-2x+2y=47 $$
$$ {{x}^{2}}+{{y}^{2}}-2x+2y-47=0 $$
Hence, it is complete solution.
Option $$(C)$$ is correct answer.

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Re-Attempt Weekly Quiz Competition

Circles Test 15

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Circles Test 15

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SHARING IS CARING

Question 1

$$2$$

$$1$$

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

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

Solution

Question 2

$${x^2} + {y^2} - 4x + 3y = 45$$

$${x^2} + {y^2} - 4x + 2y - 45 = 0$$

$${x^2} + {y^2} + 4x - 2y = 45$$

$${x^2} + {y^2} - 4x + 2y + 45 = 0$$

Solution

Question 3

Are the vertices of a right triangle

The vertices of an isosceles triangle which is not regular

Vertices of a regular triangle

Are collinear

Solution

Question 4

$$(a+b,c+d)$$

$$9-a-b,-c-d)$$

$$\left(\dfrac{a+b}{2},\dfrac{c+d}{2}\right)$$

none of these

Solution

Question 5

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

$${ x }^{ 2 }+{ y }^{ 2 }\pm 2ya=0$$

$${ x }^{ 2 }+{ y }^{ 2 }\pm 2xa=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+ya=0$$

Solution

Question 6

$$x^{2}+y^{2}+13y=0$$

$$3x^{2}+3y^{2}+13x+3=0$$

$$6x^{2}+6y^{2}-13y=0$$

$$x^{2}+y^{2}+13+3=0$$

Solution

Question 7

first quadrant

second quadrant

third quadrant

none of these

Solution

Question 8

$$2\sqrt {{r^2} - {b^2} + {a^2}} $$

$$\sqrt {{r^2} - {a^2} + {b^2}} $$

$$\left( {{r^2} - {b^2} + {a^2}} \right)$$

$$2\sqrt {{r^2} - {a^2} + {b^2}} $$

Solution

Question 9

$${ x }^{ 2 }+{ y }^{ 2 }-8x-2y-51=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+8x-2y-51=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+8x+2y-51=0$$

None of these

Solution

Question 10

$$ x^2+y^2+2x-2y-62=0$$

$$ x^2+y^2-2x+2y-62=0$$

$$ x^2+y^2-2x+2y-47=0$$

$$ x^2+y^2+2x-2y-47=0$$

Solution

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