Circles Test 27

Result

Circles Test 27

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Question 1
1 / -0

ABCD is a square of side 1 unit. A circle passes through vertices A,B of the square and the remaining two vertices of the square lie out side the circle. The length of the tangent draw to the circle from vertex D is 2 units. The radius of the circle is

A
$$\sqrt{5}$$

B
$$\dfrac{1}{2}$$$$\sqrt{10}$$

C
$$\dfrac{1}{3}$$$$\sqrt{12}$$

D
$$\sqrt{8}$$

Solution
Question 2
1 / -0

Find the equation of the circle having $$(1, -2)$$ as its centre and passing through the intersection of the lines $$3x+y=14$$ and $$2x+5y=18$$.

A
$$x^2-y^2+2x+4y+20=0$$

B
$$x^2+y^2+2x-4y+20=0$$

C
$$x^2+y^2-2x+4y-20=0$$

D
$$x^2-y^2+2x-4y+20=0$$

Solution
Question 3
1 / -0

The equation of the circle which bisects the circumference of the circles $$x^{2}+y^{2} =1, \;x^{2}+y^{2}-2x =3$$ and $$x^2+y^{2}+2y =3$$ is

A
$$x^{2}+y^{2}-4x+4y-1= 0$$

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

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

D
$$x^{2}+y^{2}-2x+4y-1= 0$$

Solution
Question 4
1 / -0

The equation of the circle passing through $$(4,6)$$ and having centre $$(1,2)$$ is

A
$${x}^{2}+{y}^{2}-2x-4y-20=0$$

B
$${x}^{2}+{y}^{2}-2x+4y-20=0$$

C
$${x}^{2}+{y}^{2}+2x-4y-20=0$$

D
$${x}^{2}+{y}^{2}+2x+4y-20=0$$

Solution

Let coordinates of point A are $$\left( 4,6 \right) $$
As circle is passing through point A, point A lies on circle.

Let center of circle is $$C\left( 1,2 \right) $$
$$\therefore h=1$$ and $$k=2$$

Thus, AC is radius of circle.

By distance formula,
$$AC=r=\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( 6-2 \right) }^{ 2 } } $$

$$\therefore r=\sqrt { { \left( 3 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 } } $$

$$\therefore r=\sqrt { 9+16 } $$

$$\therefore r=\sqrt { 25 } $$

$$\therefore r=5$$

Thus, equation of circle is,
$${ \left( x-h \right) }^{ 2 }+{ \left( y-k \right) }^{ 2 }={ r }^{ 2 }$$

$${ \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }=\left( 5 \right) ^{ 2 }$$

$${ x }^{ 2 }-2x+1+y^{ 2 }-4y+4=25$$

$$\therefore { x }^{ 2 }+y^{ 2 }-2x-4y-20=0$$

Thus, answer is option (A)
Question 5
1 / -0

The equation $$y = \sqrt{2 - x^2 - y^2} + \sqrt{x^2 + y^2 - 2}$$ represents, where $$x > 0$$

A
Circle with radius $$2$$ units

B
Parabola

C
A point circle

D
Ellipse

Solution
Question 6
1 / -0

The equation of the circle, passing through the point $$\left(2,8\right)$$, touching the lines $$4x-3y-24=0$$ and $$4x+3y-42=0$$ and having x coordinate of the centre of the circle numerically less then or equal to 8 is

A
$$x^2+y^2+4x-6y-12=0$$

B
$$x^2+y^2-4x+6y-12=0$$

C
$$x^2+y^2-4x-6y-12=0$$

D
None of these

Solution

Since $$C\left(\alpha,\beta\right)$$ be the centre of the circle.
Since the circle passes through the point $$\left(2,8\right)$$
$$\therefore$$ Radius of the circle$$=\sqrt{{\left(\alpha-2\right)}^{2}+{\left(\beta-8\right)}^{2}}$$
Since the circle touches the lines $$4x-3y-24=0$$ and $$4x+3y-42=0$$
$$\therefore\left|\dfrac{4\alpha-3\beta-24}{\sqrt{{4}^{2}+{5}^{2}}}\right|=\left|\dfrac{4\alpha+3\beta-42}{\sqrt{{4}^{2}+{5}^{2}}}\right|=\sqrt{{\left(\alpha-2\right)}^{2}+{\left(\beta-8\right)}^{2}}$$ ........$$\left(1\right)$$
Solving them, we get
$$4\alpha-3\beta-24=\pm\left(4\alpha+3\beta-42\right)$$ ........$$\left(3\right)$$
$$\therefore 6\beta=18$$ or $$\beta=3$$ by taking the positive sign
and $$8\alpha=66$$ or $$\alpha=\dfrac{66}{8}=\dfrac{33}{4}$$ by taking negative sign.
Given $$\left|\alpha\right|\le 8\Rightarrow -8\le \alpha \le 8$$
$$\therefore \alpha \neq \dfrac{33}{4}$$
Put $$\beta=3$$ in equations $$\left(1\right)$$ and $$\left(3\right)$$ and equating, we get
$${\left(4\alpha-33\right)}^{2}={\left(\alpha-2\right)}^{2}+25$$
$$\Rightarrow 16{\alpha}^{2}-264\alpha+1089=25{\alpha}^{2}+725-100\alpha$$
$$\Rightarrow 9{\alpha}^{2}+164\alpha-364=0$$ which is quadratic in $$\alpha$$
$$\therefore \alpha=\dfrac{-164\pm\sqrt{{164}^{2}-36\times -364}}{2\times 9}=\dfrac{-164\pm\sqrt{{164}^{2}+36\times 364}}{18}$$
$$=\dfrac{-164\pm\sqrt{40000}}{18}=\dfrac{-164\pm 200}{18}=2,\dfrac{-182}{9}$$
But $$-8\le \alpha\le 8$$ $$\therefore \alpha=2$$
Now $${r}^{2}={\left(\alpha-2\right)}^{2}+{\left(3-8\right)}^{2}={\left(2-2\right)}^{2}+{\left(3-8\right)}^{2}=25$$ where $$r$$ is the radius of the circle
Hence the required equation of the required circle is
$${\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}=25$$
$${x}^{2}+{y}^{2}-4x-6y+4+9-25=0$$ or $${x}^{2}+{y}^{2}-4x-6y-12=0$$
Question 7
1 / -0

The centre of family of circles cutting the family of circles $$x^2+y^2+4x(\lambda-\dfrac{3}{2})+3y(\lambda-\dfrac{4}{3})-6(\lambda+2)=0$$, lies on

A
$$x-y-1=0$$

B
$$4x+3y-6=0$$

C
$$4x+3y+7=0$$

D
$$3x-4y-1=0$$
Question 8
1 / -0

If $$x=2+3\cos\theta$$ and $$y=1-3\sin\theta$$ represent a circle then the centre and radius is?

A
$$(2, 1), 9$$

B
$$(2, 1), 3$$

C
$$(1, 2), \dfrac{1}{3}$$

D
$$(-2, -1), 3$$

Solution

We have,
$$x=2+3\cos\theta\Rightarrow 3\cos \theta\Rightarrow 3\cos\theta =x-2$$ ..(i)
$$y=1-3\sin\theta \Rightarrow 3\sin\theta =-y+1$$ ..(ii)
Squaring adding (i) & (ii), we get
$$(x-2)^2(-(y-1))^2=3^2(\cos^2\theta +\sin^2\theta)$$
$$\Rightarrow (x-2)^2+(-(y-1))^2=3^2$$
$$\therefore$$ Radius $$=3$$, centre $$=(2, 1)$$.
Question 9
1 / -0

The equation $${ x }^{ 2 }+{ y }^{ 2 }+4x+6y+13=0\quad $$ represents

A
Circle

B
Pair of coincident straight lines

C
Pair of concurrent straight lines

D
Point

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
$$a \in ( - \infty , - 1 )$$

B
$$a \epsilon ( - 1,2 )$$

C
$$a \epsilon ( 2 , \infty )$$

D
$$a \epsilon ( - \infty , - 1 ) \cup ( 2 , \infty )$$

Solution