Circle MCQ Questions & Answers in Geometry | Maths

The given circle is $${x^2} + {y^2} - 2x - 6y + 6 = 0$$      with centre $$C\left( {1,\,3} \right)$$   and radius $$ = \sqrt {1 + 9 - 6} = 2$$
Let $$AB$$  be one of its diameter which is the chord of other circle with centre at $${C_1}\left( {2,\,1} \right)$$
Then in $$\Delta {C_1}CB,$$
$$\eqalign{ & {C_1}{B^2} = CC_1^2 + C{B^2} \cr & {r^2} = \left[ {{{\left( {2 - 1} \right)}^2} + {{\left( {1 - 3} \right)}^2}} \right] + {\left( 2 \right)^2} \cr & \Rightarrow {r^2} = 1 + 4 + 4 \cr & \Rightarrow {r^2} = 9 \cr & \Rightarrow r = 3 \cr} $$

The common chord is $${x^2} + {\left( {y - \lambda } \right)^2} - 16 - \left\{ {{x^2} + {y^2} - 16} \right\} = 0,$$         i.e., $$y = \frac{\lambda }{2}\,\,\,\,\,\left( {\because \,\,\lambda \ne 0} \right)$$
The pair of straight lines joining the origin to the points of intersection of $$y = \frac{\lambda }{2}$$   and $${x^2} + {y^2} = 16$$   is $${x^2} + {y^2} = 16.{\left( {\frac{{2y}}{\lambda }} \right)^2},$$     i.e., $${\lambda ^2}{x^2} + \left( {{\lambda ^2} - 64} \right){y^2} = 0$$
These lines are at right angles if $${\lambda ^2} + {\lambda ^2} - 64 = 0$$    i.e., $$\lambda = \pm 4\sqrt 2.$$

Polar of the circle is $$xx' + yy' = {a^2},$$    but it given by
$$Ax + By + C = 0,$$    then $$\frac{{x'}}{A} = \frac{{y'}}{B} = \frac{{{a^2}}}{{ - C}}$$
Hence pole is $$\left( {\frac{{{a^2}A}}{{ - C}},\,\frac{{{a^2}B}}{{ - C}}} \right)$$

Any line through $$P\left( { - 2,\, - 1} \right)$$   is $$y + 1 = m\left( {x + 2} \right)$$
It touches the circle if $$\left| {\frac{{2m - 1}}{{\sqrt {1 + {m^2}} }}} \right| = 1,$$    i.e., $$m = 0,\,\frac{4}{3}$$
$$\therefore $$  the equation of $$PB$$  is $$y + 1 = \frac{4}{3}\left( {x + 2} \right),$$    i.e., $$4x - 3y + 5 = 0$$
A point on $$PB$$  is $$\left( { - 5,\, - 5} \right).$$   Its image by the line $$y = - 1$$   is $$P'\left( { - 5,\,3} \right).$$   So, the equation of the incident ray $$P'P$$  is
$$y - 3 = \frac{{3 + 1}}{{ - 5 + 2}}\left( {x + 5} \right){\text{ or }}4x + 3y + 11 = 0$$

Solving $$y = x,\,{x^2} + {y^2} - 2x = 0,$$     we get $$\left( {x,\,y} \right) = \left( {0,\,0} \right),\,\left( {1,\,1} \right).$$     So, the ends of the diameter are $$\left( {0,\,0} \right),\,\left( {1,\,1} \right).$$   Hence, the equation of the required circle is $$\left( {x - 0} \right)\left( {x - 1} \right) + \left( {y - 0} \right)\left( {y - 1} \right) = 0.$$

The points of intersection of $$2x - y + 1 = 0$$    with the axes are $$\left( { - \frac{1}{2},\,0} \right)$$  and $$\left( {0,\,1} \right).$$
The points of intersection of $$x + \lambda y - 3 = 0$$    and the axes are $$\left( {3,\,0} \right),\,\left( {0,\,\frac{3}{\lambda }} \right).$$
The points are concyclic if $$OA.OC = OB.OD$$
$$ \Rightarrow \frac{1}{2}.3 = 1.\left| {\frac{3}{\lambda }} \right|{\text{ or }}\left| \lambda \right| = 2$$
As $$\lambda $$ must be negative according to the geometrical situation, $$\lambda = - 2.$$

$$\pi {r^2} = 154 \Rightarrow r = 7$$
For centre on solving equation
$$2x - 3y = 5\,\,\& \,\,3x - 4y = 7$$       we get $$x=1,y=-1$$
$$\therefore $$ centre $$ = \left( {1,\, - 1} \right)$$
Equation of circle,
$$\eqalign{ & {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2} \cr & {x^2} + {y^2} - 2x + 2y = 47 \cr} $$

$${x^2} + {y^2} + \frac{\lambda }{2}x - \frac{{1 + {\lambda ^2}}}{2}y - 5 = 0$$       and $${x^2} + {y^2} + 4x + 6y + 3 = 0$$      cut orthogonally if $$2.\frac{\lambda }{4}.2 + 2.\left( { - \frac{{1 + {\lambda ^2}}}{4}} \right).3 = - 5 + 3 \Rightarrow $$         two real values of $$\lambda .$$

Let there be a circle of radius $$R$$ and $$AB$$  a chord.
$$\eqalign{ & OD \bot AB{\text{ and }}AD = DB \cr & {\text{and }}AB = 2AD \cr & \angle AOB = \theta \cr & \Rightarrow \angle AOD = \frac{\theta }{2} \cr & {\text{In }}\Delta AOD, \cr & \sin \,\frac{\theta }{2} = \frac{{AD}}{{OA}} \cr & \sin \,\frac{\theta }{2} = \frac{{AD}}{R} \cr & AD = R\,\sin \,\frac{\theta }{2} \cr} $$
$$\therefore $$  Length of chord $$AB = 2AD = 2R\,\sin \frac{\theta }{2}$$

$${x^2} + {y^2} = {r^2}$$     is a circle with centre at $$\left( {0,\,0} \right)$$  and radius $$r$$ units.

Any arbitrary point $$P$$ on it is $$\left( {r\,\cos \,\theta ,\,r\,\sin \,\theta } \right)$$
Choosing $$A$$ and $$B$$ as $$\left( { - r,\,0} \right)$$  and $$\left( { 0,\,- r} \right)$$
[So that $$\angle AOB = {90^ \circ }$$   ]
For locus of centroid of $$\Delta ABP$$
$$\eqalign{ & \left( {\frac{{r\,\cos \,\theta - r}}{3},\,\frac{{r\,\sin \,\theta - r}}{3}} \right) = \left( {x,\,y} \right) \cr & \Rightarrow r\,\cos \,\theta - r = 3x \cr & \,\,\,\,\,\,r\,\sin \,\theta - r = 3y \cr & \Rightarrow r\,\cos \,\theta = 3x + r \cr & \,\,\,\,\,\,r\,\sin \,\theta = 3y + r \cr} $$
Squaring and adding $${\left( {3x + r} \right)^2} + {\left( {3y + r} \right)^2} = {r^2}$$      which is a circle.