Circles Test 13

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Circles Test 13

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Weekly Quiz Competition

Question 1
1 / -0

Find the equation of the circle which touches the coordinate axes and whose centre lies on the line $$x - 2y = 3$$.

A
$$(x-1)^2+(y+1)^2=1$$

B
$$(x+1)^2+(y+1)^2=1$$

C
$$(x-1)^2+(y-1)^2=1$$

D
$$(x+1)^2+(y-1)^2=1$$

Solution

Given:
The equation of line is $$ x - 2y = 3 $$
Substitute $$x = 0$$ in the given equation.

$$ 0 - 2y = 3 \Rightarrow - 2y = 3 \Rightarrow y = - \dfrac{3}{2}$$

Substitute $$y = 0$$ in the given equation.

$$ x - 2\left (0 \right ) = 3 \Rightarrow x = 3 $$

The points are $$ \left (3, 0 \right ) \, and \, \left (0, - \dfrac{3}{2} \right ) $$.

Let the centre be $$ \left (a, -a \right ) $$.

Substitute $$ \left (a, -a \right ) $$ in given equation.

$$ a + 2a = 3 \Rightarrow 3a = 3 \Rightarrow a = 1 $$

Centre $$ = \left (1, -1 \right ) $$

Radius $$ = 1$$

The equation of a circle becomes $$ \left (x - 1 \right )^{2} + \left (y + 1 \right )^{2} = 1 $$.
Hence, the required solution is found.
Question 2
1 / -0

If $$A(5,-4)$$ and $$B(7,6)$$ are points in a plane, then the set of all points $$P(x,y)$$ in the plane such that $$AP=PB=2:3$$ is

A
a circle

B
a hyperbola

C
an ellipse

D
a parabola
Question 3
1 / -0

The equation of the circle which passes through the points $$(2, 3)$$ and $$(4. 5)$$ and the centre lies on the straight line $$y - 4x + 3 = 0$$, is

A
$$x^{2} + y^{2} + 4x - 10y + 25 = 0$$

B
$$x^{2} - y^{2} - 4x - 10y + 25 = 0$$

C
$$x^{2} + y^{2} - 4x - 10y + 16 = 0$$

D
$$x^{2} + y^{2} - 14y + 8 = 0$$

Solution

Centre lies on the line $$y-4x+3=0$$

Let $$x=h$$

$$\Rightarrow y=4h-3$$

So the center is of the form $$(h,4h-3)$$

Distance of centre from $$(2,3)$$ and $$(4,5)$$ will be equal

$$\Rightarrow { (h-2) }^{ 2 }+{ (4h-3-3) }^{ 2 }={ (h-4) }^{ 2 }+{ (4h-3-5) }^{ 2 }\\ \Rightarrow h=2$$

So the centre is $$(2,5)$$

$$r=\sqrt { { (2-2) }^{ 2 }+{ (5-3) }^{ 2 } } =2$$

So the equation of the circle is

$${ (x-2) }^{ 2 }+{ (y-5) }^{ 2 }={ 2 }^{ 2 }\\ { x }^{ 2 }+{ y }^{ 2 }-4x-10y+25=0$$
Question 4
1 / -0

The equation of the image of the circle $${ x }^{ 2 }+{ y }^{ 2 }+16x-24y+183=0$$ by the line mirror $$4x+7y+13=0$$ is:

A
$${ x }^{ 2 }+{ y }^{ 2 }+32x-4y+235=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }+32x+4y-235=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }+32x-4y-235=0$$

D
$${ x }^{ 2 }+{ y }^{ 2 }+32x+4y+235=0$$

Solution

Equation of given circle:
$$x^2+y^2++16x-24y+183=0$$

$$\textbf{Idea}$$; Lets find the reflection of centre of this circle with respect to the given line equation. Then find the radius of given circle. After reflection also, the radius of circle does not change. Thus finally knowing the centre of reflected circle and its radius we can find equation of reflected circle.

First convert this equation to standard form of circle equation as follows:
$$x^2+y^2++16x-24y+183=0$$
$$(x^2+16x+64)+(y^2-24y+144)-25=0$$
$$(x+8)^2+(y-12)^2-25=0$$
$$(x+8)^2+(y-12)^2=5^2$$
Thus $$\text{Centre}= (-8,12)$$
and $$\text{radius}=\sqrt{25}=5$$

Let reflected centre have co-ordinates= $$(h,k)$$
Now note that mid point of actual centre and reflected centre will definitely fall on line of reflection.
Mid-point of $$(h,k)$$ and $$(-8,12)$$=$$\left (\dfrac{h-8}{2},\dfrac{k+12}{2}\right)$$
Lets try to satisfy line equation with given point.
$$4x+7y+13=0\rightarrow4\left (\dfrac{h-8}{2}\right)+7\left (\dfrac{k+12}{2}\right)+13=0$$
$$4h+7k+78=0$$ .....(1)

Also slope of given line is perpendicular to slope of RC
Slope of given line$$=-4/7$$
$$m_1 m_2=-1$$
$$\dfrac{\dfrac{(k+12)}{2}-12} { \left (\dfrac{h-8}{2}\right) +8}\times\dfrac{-4}{7}=-1$$
$$\rightarrow7h-4k+104=0$$ ...........(2)
Solving (1) and (2) we get, $$(h,k)=(-16,-2)$$
Equation of circle with centre $$(-16,-2)$$ and radius $$=5$$:
$$(x+16)^2+(y+2)^2=5^2$$
$$x^2+y^2+32x+4y+235=0$$
Question 5
1 / -0

The equation of the circle, which is the mirror image of the circle, $${ x }^{ 2 }+{ y }^{ 2 }-2x=0$$, in the line, $$y=3-x$$ is:

A
$${ x }^{ 2 }+{ y }^{ 2 }-6x-8y+24=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }-8x-6y+24=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }-4x-6y+12=0\quad $$

D
$${ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12=0\quad $$

Solution

Centre of circle $$(1,0)$$, radius $$= 1$$
Image of $$(1,0)$$ w.r.t. Line $$x+y-3=0$$ is
$$\dfrac{x-1}{1} = \dfrac{y-0}{1} = \dfrac{-2(1+0-3)}{1+1}=2$$
$$\Rightarrow x=3, \,\,y=2$$
Now circle with centre $$(3,2)$$ and radius $$1$$ is $$(x-3)^2+(y-2)^2=1$$
$$\Rightarrow x^2+9-6x+y^2+4-4y=1$$
$$\Rightarrow x^2+y^2-6x-4y+12=0$$
Question 6
1 / -0

Point $$\left( 0,\lambda \right) $$ lies in the interior of circle $$x^2+y^2=c^2$$ then

A
$$\lambda \in \left( -c,c\right) $$

B
$$\lambda \in \left(0,c \right) $$

C
$$\lambda \in \left(-c,0\right) $$

D
None.

Solution

The given equation is $$x^2+y^2=c^2$$
Condition of a point to be interior of circle
$$\implies x^2+y^2<c^2\\\implies 0+\lambda ^2<c^2\\\lambda ^2<c^2\\\lambda\in(-c,c)$$
Question 7
1 / -0

State whether following statements are true or false
Statement-1 : The only circle having radius $$\sqrt {10}$$ and a diameter along line $$2x + y=5$$ is $$x^2 + y^2 - 6x + 2y=0$$.
Statement-2: The line 2x + y=5 is a normal to the circle $$x^2+ y^2 - 6x + 2y =0.$$

A
Statement- 1 is false, statement-2 is true.

B
Statement-1 is true,statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

C
Statement-1 is true, statement-2 is false.

D
Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Solution

Statement 1 and 2 both are correct, but statement 2 is not correct explaination for statement 1.

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

$$(-g,-f)=(-3,1), c=0$$

$$r^2=g^2+f^2-c$$

$$\therefore r^2=9+1-0=10$$

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

$$2x+y=5$$ is also a normal to the circle.
Question 8
1 / -0

The equation of the circle passing through $$(4,\ 5)$$ having the centre at $$(2 ,\ 2)$$ is

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

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

C
$$x^{2} + y^{2} -4x = 13$$

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

Solution

Let the equation of circle be
$$x^{2}+y^{2}+2gx+2fy+c=0$$
$$\therefore$$ as centre is $$(-g, -f)$$
$$g=-2, f=-2$$ ($$\therefore$$ as centre if given $$(2,2)$$)
$$\therefore$$ eqn is $$x^{2}+y^{2}-4x-4y+c=0$$
Also circle passes through $$(4,5)$$
$$\therefore$$ it satisfies $$(4,5)$$
$$\therefore (4)^{2}+(5)^{2}-4(4)-4(5)+C=0$$
$$\therefore 16+25-16-20+C=0$$
$$\therefore 5+C=0$$
$$\therefore C=-5$$
Hence eqn of circle is
$$x^{2}+y^{2}-4x-4y-5=0$$
Question 9
1 / -0

Find the equation of the circle which passes through the point $$(1, 1)$$ & which touches the $$x^2 + y^2 + 4x - 6y - 3 = 0$$ at the point $$(2, 3)$$ on it.

A
$$x^2 + y^2 + x - 6y + 3 = 0$$

B
$$x^2 + y^2 + x - 6y - 3 = 0$$

C
$$x^2 + y^2 + x + 6y + 3 = 0$$

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

Solution

Correction(in 3rd last step): $$x^2+y^2 ``+"x-6y=\dfrac{9}{4}-\dfrac{1}{4}+4-9$$
Change $$-x$$ to $$+x$$ in following steps
Question 10
1 / -0

Find the equation of the circle which passes through the point (1, 1) & which touches the circle $$x^{2} + y^{2} + 4x - 6y - 3 = 0$$ at the point $$(2, 3)$$ on it.

A
$$x^{2} + y^{2} + x - 6y + 3 = 0$$

B
$$x^{2} + y^{2} + x - 6y - 3 = 0$$

C
$$x^{2} + y^{2} + x + 6y + 3 = 0$$

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

Solution

REF.Image.
Let's consider given circle

$$ S_{1} = 0 $$ &

equation of required circle

$$as\, S_{2} =0 $$

$$ Eq^n $$ of tangent at P(2,3) to $$ S_{1} = 0 $$

is $$ x = 2 $$

(because equation of CP is y = 3 and tangent at P is $$ \perp ^{le}$$ to
Required circle should touch $$ S_{1} =0 $$ at $$P(2,3)$$

$$ \Rightarrow $$ tangent at $$ P(2,3) $$ to $$ S_{1} = 0$$ should be
common tangent for $$ S_{1} = 0 \, \&\, S_{2} = 0$$

$$ \Rightarrow $$ Required circle should touch the line x = 2
at P(2,3)

$$ eq^n $$ of $$ S_{2} =0 $$ will belong to family of circles
$$ S+\lambda L = 0$$ [ S is point circle of L is tangent at p]
i.e.$$ (x-2)^2 + (y-3)^2 +\lambda (x-2) = 0$$

It should satisfy (1,1) $$ \Rightarrow (1-2)^2+(1-3)^2+ \lambda (1-2) = 0$$
$$ \Rightarrow \lambda = 5 $$

equation of required circle
$$ (x-2)^2 + (y-3)^2 + 5(x-2) = 0$$
$$ \Rightarrow x^2 +y^2 +x -6y +3 =0 $$

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Re-Attempt Weekly Quiz Competition

Circles Test 13

Result

Circles Test 13

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SHARING IS CARING

Question 1

$$(x-1)^2+(y+1)^2=1$$

$$(x+1)^2+(y+1)^2=1$$

$$(x-1)^2+(y-1)^2=1$$

$$(x+1)^2+(y-1)^2=1$$

Solution

Question 2

a circle

a hyperbola

an ellipse

a parabola

Question 3

$$x^{2} + y^{2} + 4x - 10y + 25 = 0$$

$$x^{2} - y^{2} - 4x - 10y + 25 = 0$$

$$x^{2} + y^{2} - 4x - 10y + 16 = 0$$

$$x^{2} + y^{2} - 14y + 8 = 0$$

Solution

Question 4

$${ x }^{ 2 }+{ y }^{ 2 }+32x-4y+235=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+32x+4y-235=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+32x-4y-235=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+32x+4y+235=0$$

Solution

Question 5

$${ x }^{ 2 }+{ y }^{ 2 }-6x-8y+24=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-8x-6y+24=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-4x-6y+12=0\quad $$

$${ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12=0\quad $$

Solution

Question 6

$$\lambda \in \left( -c,c\right) $$

$$\lambda \in \left(0,c \right) $$

$$\lambda \in \left(-c,0\right) $$

None.

Solution

Question 7

Statement- 1 is false, statement-2 is true.

Statement-1 is true,statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

Statement-1 is true, statement-2 is false.

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Solution

Question 8

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

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

$$x^{2} + y^{2} -4x = 13$$

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

Solution

Question 9

$$x^2 + y^2 + x - 6y + 3 = 0$$

$$x^2 + y^2 + x - 6y - 3 = 0$$

$$x^2 + y^2 + x + 6y + 3 = 0$$

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

Solution

Question 10

$$x^{2} + y^{2} + x - 6y + 3 = 0$$

$$x^{2} + y^{2} + x - 6y - 3 = 0$$

$$x^{2} + y^{2} + x + 6y + 3 = 0$$

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

Solution

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