Circle MCQ Questions & Answers in Geometry | Maths

Given line is $$y = mx + c.....\left( 1 \right)$$
and the given circle is $${x^2} + {y^2} = {r^2}......\left( 2 \right)$$
Solving equations $$\left( 1 \right)$$ and $$\left( 2 \right),$$ we get
$$\left( {1 + {m^2}} \right){x^2} + 2mcx + {c^2} - {r^2} = 0......\left( 3 \right)$$
For two real distinct points of intersection, both the roots of $$\left( 3 \right)$$ must be real and distinct.
$$\eqalign{ & \therefore \,4{m^2}{c^2} - 4\left( {1 + {m^2}} \right)\left( {{c^2} - {r^2}} \right) > 0 \cr & \Rightarrow {c^2} < {r^2}\left( {1 + {m^2}} \right) \cr & \Rightarrow - r\sqrt {1 + {m^2}} < c < r\sqrt {1 + {m^2}} \cr} $$

The centre of the circle is $$C\left( {1,\,1} \right)$$   and radius of the circle is $$2,$$ perpendicular distance from $$C$$ on $$AB,$$  the chord $$x + y = 3$$

$$\eqalign{ & CD = \left| {\frac{{1 + 1 - 3}}{{\sqrt 2 }}} \right| = \frac{1}{{\sqrt 2 }} \cr & \therefore \,AD = \sqrt {4 - \frac{1}{2}} = \sqrt {\frac{7}{2}} \,\,\,\,\,\,\,\,\,\left[ {AD = \sqrt {A{C^2} - C{D^2}} } \right] \cr} $$
Hence, the length of the chord $$AB = 2AD = 2\sqrt {\frac{7}{2}} = \sqrt {14} $$

$$\eqalign{ & {x^2} - 8x + 12 = 0\,\,\, \Rightarrow \left( {x - 6} \right)\left( {x - 2} \right) = 0 \cr & {y^2} - 14y + 45 = 0\,\,\, \Rightarrow \left( {y - 5} \right)\left( {y - 9} \right) = 0 \cr} $$
Thus sides of square are
$$x = 2,\,\,x = 6,\,\,y = 5,\,\,y = 9$$
Then centre of circle inscribed in square will be
$$\left( {\frac{{2 + 6}}{2},\,\frac{{5 + 9}}{2}} \right) = \left( {4,\,7} \right)$$

As $$2x-3y-5=0$$    and $$3x-4y-7=0$$    are diameters of circles.
$$\therefore $$ Centre of circle is solution of these two equation $$'s,$$ i.e.
$$\eqalign{ & \frac{x}{{21 - 20}} = \frac{y}{{ - 15 + 14}} = \frac{1}{{ - 8 + 9}} \cr & \Rightarrow x = 1,\,\,y = - 1 \cr & \therefore C\left( {1,\, - 1} \right) \cr} $$
Also area of circle, $$\pi {r^2} = 154$$
$$\eqalign{ & \Rightarrow {r^2} = \frac{{154}}{{22}} \times 7 = 49 \cr & \Rightarrow r = 7 \cr} $$
$$\therefore $$ Equation of required circle is
$$\eqalign{ & {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2} \cr & \Rightarrow {x^2} + {y^2} - 2x + 2y = 47 \cr} $$

Let the required circle be $${x^2} + {y^2} + 2gx + 2fy + c = 0$$
Since it passes through $$\left( {0,\,0} \right)$$  and $$\left( {1,\,0} \right)$$
$$ \Rightarrow c = 0{\text{ and }}g = - \frac{1}{2}$$
Points $$\left( {0,\,0} \right)$$  and $$\left( {1,\,0} \right)$$  lie inside the circle $${x^2} + {y^2} = 9,$$   so two circles touch internally
$$\eqalign{ & \Rightarrow {c_1}{c_2} = {r_1} - {r_2} \cr & \therefore \sqrt {{g^2} + {f^2}} = 3 - \sqrt {{g^2} + {f^2}} \cr & \Rightarrow \sqrt {{g^2} + {f^2}} = \frac{3}{2} \cr & \Rightarrow {f^2} = \frac{9}{4} - \frac{1}{4} = 2 \cr & \therefore f = \pm \sqrt 2 \cr} $$
Hence, the centres of required circle are
$$\left( {\frac{1}{2},\,\sqrt 2 } \right){\text{ or }}\left( {\frac{1}{2},\, - \sqrt 2 } \right)$$

Centre and the radius of the given circle are $$C\left( {5,\,7} \right)$$  and $$r = \sqrt {25 + 49 + 51} = 5\sqrt 5 $$      respectively.
$$\eqalign{ & {\text{Let }}P = \left( { - 7,\,2} \right){\text{ be the given point}}{\text{.}} \cr & {\text{Now }}CP = \sqrt {{{12}^2} + {5^2}} = 13 \cr & {\text{Thus }}p = CP + r = 13 + 5\sqrt 5 {\text{ and}} \cr & q = CP - r = 13 - 5\sqrt 5 \cr & {\text{Hence G}}{\text{.M}}{\text{. of }}p{\text{ and }}q{\text{ is}} \cr & = \sqrt {pq} = \sqrt {{{13}^2} + 125} = 2\sqrt {11} \cr} $$

The circle through points of intersection of the two circles $${x^2} + {y^2} - 6 = 0$$    and $${x^2} + {y^2} - 6x + 8 = 0$$     is $$\left( {{x^2} + {y^2} - 6} \right) + \lambda \left( {{x^2} + {y^2} - 6x + 8} \right) = 0$$
As it passes through (1, 1)
$$\left( {1 + 1 - 6} \right) + \left( {1 + 1 - 6 + 8} \right) = 0 \Rightarrow \lambda = \frac{4}{4} = 1$$
$$\therefore $$ The required circle is
$$2{x^2} + 2{y^2} - 6x + 2 = 0\,\,\,{\text{or,}}\,\,{x^2} + {y^2} - 3x + 1 = 0$$

The number of points is equal to the number of integral solutions $$\left( {x,\,y} \right)$$  such that $${x^2} + {y^2} < 16.$$   So $$x,\,y$$  are integers such that $$ - 3\, \leqslant x \leqslant 3,\, - 3 \leqslant y \leqslant 3$$      satisfying the inequation $${x^2} + {y^2} < 16.$$   The number of selections of values of $$x$$ is 7, namely $$ - 3,\, - 2,\, - 1,\,0,\,1,\,2,\,3.$$     The same is true for $$y.$$ So the number of ordered pairs $$\left( {x,\,y} \right)$$  is $$7 \times 7.$$  But $$\left( {3,\,3} \right),\,\left( {3,\, - 3} \right),\,\left( { - 3,\,3} \right),\,\left( { - 3,\, - 3} \right)$$       are rejected because they do not satisfy the inequation $${x^2} + {y^2} < 16.$$   So, the number of points is 45.

Here, $$OB =$$  radius $$= 2.$$
The distance of (0, 0) from $$x + y = 5\sqrt 2 $$   is 5.
$$\therefore $$  the radius of the smallest circle $$ = \frac{{5 - 2}}{2} = \frac{3}{2}$$   and $$OC = 2 + \frac{3}{2} = \frac{7}{2}$$
The slope of $$OA = 1 = \tan \,\theta $$
$$\eqalign{ & \therefore \,\cos \,\theta = \frac{1}{{\sqrt 2 }},\,\,\sin \,\theta = \frac{1}{{\sqrt 2 }} \cr & \therefore \,C = \left( {0 + OC.\cos \,\theta ,\,0 + OC.\sin \,\theta } \right) = \left( {\frac{7}{{2\sqrt 2 }},\,\frac{7}{{2\sqrt 2 }}} \right) \cr} $$

Equation of circle whose centre is $$\left( {h,\,k} \right)$$
i.e., $${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {k^2}$$

(radius of circle $$=k$$  because circle is tangent to $$x$$-axis)
Equation of circle passing through $$\left( { - 1,\, + 1} \right)$$
$$\eqalign{ & \therefore {\left( { - 1 - h} \right)^2} + {\left( {1 - k} \right)^2} = {k^2} \cr & \Rightarrow 1 + {h^2} + 2h + 1 + {k^2} - 2k = {k^2} \cr & \Rightarrow {h^2} + 2h - 2k + 2 = 0 \cr & D \geqslant 0 \cr & \therefore {\left( 2 \right)^2} - 4 \times 1.\left( { - 2k + 2} \right) \geqslant 0 \cr & \Rightarrow 4 - 4\left( { - 2k + 2} \right) \geqslant 0 \cr & \Rightarrow 1 + 2k - 2 \geqslant 0 \cr & \Rightarrow k \geqslant \frac{1}{2} \cr} $$