Circles Test 17

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Circles Test 17

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Weekly Quiz Competition

Question 1
1 / -0

$$P$$ and $$Q$$ are any two points on the circle $$x^2 + y^2 = 4$$ such that $$PQ$$ is a diameter. If $$\alpha$$ and $$\beta$$ are the lengths of perpendicular from $$P$$ and $$Q$$ on $$x + y = 1$$ then the maximum value of $$\alpha \beta$$ is

A
$$\dfrac{1}{2}$$

B
$$\dfrac{7}{2}$$

C
$$1$$

D
$$2$$

Solution
Question 2
1 / -0

Two rods of lengths 'a' and 'b' slide along coordinate aces such that their ends are concyclic. Locus of the centre of the circle is?

A
$$4(x^2+y^2)=a^2+b^2$$

B
$$4x(x^2+y^2)=a^2-b^2$$

C
$$4(x^2-y^2)=a^2-b^2$$

D
$$xy=ab$$

Solution

$$k^{2}+ \left( \dfrac{b}{\alpha} \right)^{2} =r^{2}$$
$$h^{2}+ \left( \dfrac{a}{\alpha} \right)^{2} =r^{2}$$
$$\Rightarrow 4 (h^{2}+k^{2})=a^{2}+b^{2}$$
Question 3
1 / -0

The length of the diameter of the circle which touches the $$x-$$axis at the point $$(1,0)$$ and passes through the point $$(2,3)$$

A
$$6/5$$

B
$$5/3$$

C
$$10/3$$

D
$$3/5$$.

Solution

Given:-
A circle which touches the $$x$$-axxis at $$\left( 1, 0 \right)$$ and also passes through the point $$\left( 2, 3 \right)$$.
To find:-
Diameter of the circle
Solution:-
Let the coordinates of centre of the circle be $$\left( h, k \right)$$ and $$r$$ be the radius of the circle.
Now,
Since the circle touches $$x$$-axis at $$\left( 1, 0 \right)$$.
$$\therefore$$ Radius of circle $$\left( r \right) = k$$
Therefore, equation of circle is-
$${\left( x - h \right)}^{2} + {\left( y - k \right)}^{2} = {k}^{2}$$
Given that the circle passes through points $$\left( 1, 0 \right)$$ and $$\left( 2, 3 \right)$$, thus both coordinates will satisfy the equation of circle.
Therefore,
$${\left( 1 - h \right)}^{2} + {k}^{2} = {k}^{2}$$
$$\Rightarrow {\left( h - 1 \right)}^{2} = 0$$
$$\Rightarrow h = 1 ..... \left( 1 \right)$$
$${\left( 2 - h \right)}^{2} + {\left( 3 - k \right)}^{2} = {k}^{2}$$
$$\Rightarrow {h}^{2} + 4 - 4h + 9 + {k}^{2} - 6k = {k}^{2}$$
$$\Rightarrow {h}^{2} - 4h - 6k + 13 = 0 ..... \left( 2 \right)$$
From $${eq}^{n} \left( 1 \right)$$, we have
$${1}^{2} - 4 \left( 1 \right) - 6k + 13 = 0$$
$$1 - 4 - 6k + 13 = 0$$
$$\Rightarrow 6k = 10$$
$$\Rightarrow k = \cfrac{5}{3}$$
Therefore,
Radius of circle $$= \cfrac{5}{3}$$
$$\therefore$$ Diameter of circle $$= 2r = 2 \times \cfrac{5}{3} = \cfrac{10}{3}$$
Thus the diameter of circle is $$\cfrac{10}{3}$$.
Hence the correct answer is $$\left( C \right) \cfrac{10}{3}$$
Question 4
1 / -0

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

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

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

Equation of circle :
$${ \left( x-2 \right) }^{ 2 }+{ \left( y-\left( -1 \right) \right) }^{ 2 }={ \left( \sqrt { { \left( 3-2 \right) }^{ 2 }+{ \left( 6-\left( -1 \right) \right) }^{ 2 } } \right) }^{ 2 }$$
$$\therefore { x }^{ 2 }-4x+4+{ y }^{ 2 }+2y+1={ \left( \sqrt { 1+49 } \right) }^{ 2 }$$
$$\therefore \boxed { { x }^{ 2 }+{ y }^{ 2 }-4x+2y=45 }$$
Equation of circle.
Question 5
1 / -0

The circle passing through the points $$(-1,0)$$ and touching the y-axis at $$(0,2)$$ also passes through the point:

A
$$(-\dfrac{3}{2},0)$$

B
$$(-\dfrac{5}{2},2)$$

C
$$(-\dfrac{3}{2},\dfrac{5}{2})$$

D
$$(-4,0)$$

Solution

Let $$\left( h, k \right)$$ be center of circle.
Circle touches the $$y$$-axis.

$$\therefore$$ Radius of circle $$= h$$

Equation of circle-

$${\left( x - h \right)}^{2} + {\left( y - k \right)}^{2} = {h}^{2} ..... \left( 1 \right)$$
Since the circle passes through $$\left( 0, 2 \right)$$

Therefore,

$${h}^{2} + {\left( 2 - k \right)}^{2} = {h}^{2}$$

$$\Rightarrow {\left( k - 2 \right)}^{2} = 0$$

$$\Rightarrow k = 2$$

Given that the circle also passes through $$\left( -1, 0 \right)$$,

Therefore,

$${\left( -1 - h \right)}^{2} + {2}^{2} = {h}^{2}$$

$${h}^{2} + 1 + 2h + 4 = {h}^{2}$$

$$h = - \cfrac{5}{2}$$

Substituting the value of $$h$$ and $$k$$ in $${eq}^{n} \left( 1 \right)$$, we get

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

Now, we can find the point from which the circle passes.

As the point $$\left( 4, 0 \right)$$ satisfy the equation of circle.

Thus, the circle will also pass through the point $$\left( -4, 0 \right)$$.

Hence the correct answer is $$\left( D \right) \left( -4, 0 \right)$$.
Question 6
1 / -0

The equation of the circle having the lines $$y^{2}+2y+4x-2xy=0$$ as its normals & passing through the point$$(2,1)$$ is

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

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

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

D
$$x^{2}+y^{2}+2x+4y+13=0$$
Question 7
1 / -0

If $$(6, -3)$$ is the one extremity of diameter to the circle $$x^{2}+y^{2}-3x+8y-4=0$$ then its other extremity is-

A
$$(3/2, -4)$$

B
$$(-3, -5)$$

C
$$(3, -5)$$

D
$$(3, 5)$$

Solution
Question 8
1 / -0

If $$2x - 3y = 5$$ and $$3x - 4y = 7$$ are the equation of $$2$$ diameters of a circle whose area is $$88$$ sq. units then the equation of the circle is :

A
$${x^2} + {y^2} + 2x - 2y - 47 = 0$$

B
$${x^2} + {y^2} - 2x + 2y - 49 = 0$$

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

D
$${x^2} + {y^2} - 2x + 2y - 26 = 0$$

Solution
Question 9
1 / -0

The centres of a set of circles, each of radius $$3$$, lie on the circle $${x}^{2}+{y}^{2}=25$$. The lotus of any point in the set is

A
$$4\le {x}^{2}+{y}^{2}\le 64$$

B
$${x}^{2}+{y}^{2}\le 25$$

C
$${x}^{2}+{y}^{2}\ge 25$$

D
$$3\le {x}^{2}+{y}^{2}\le 9$$

Solution
Question 10
1 / -0

If he equations of two diameters of a circle are $$2x + y = 6$$ and $$3x + 2y = 4$$ and the radius is $$10$$ , find the equation of the circle.

A
$$x^{2}+y^{2}-16x+20y+64=0$$

B
$$x^{2}-y^{2}+16x-20y+34=0$$

C
$$x^{2}+y^{2}+6x-20y+64=0$$

D
None of these

Solution

Equation of two diameters of circle are
$$2x+y=6 ....... (i)$$
$$3x+2y=4 ...... (ii)$$

We know intersection of these line will give centre of circle

$$4x+2y=12$$
$$3x+2y=4$$
$$(-)$$ $$(-)$$
$$\overline {2x+0=16}$$

$$\Rightarrow x=8$$

Put $$x=8$$ in $$(i)$$ we get $$16+y=6$$
$$y=-10$$

centre of circle is $$(8, -10)$$ and radius $$10$$ equation of circle will be

$$(x-8)^{2}+(y+10)^{2}=(10)^{2}$$

$$x^{2}-16x+64+y^{2}+20y+100=100$$

$$x^{2}+y^{2}-16x+20y+64=0$$