Circle MCQ Questions & Answers in Geometry | Maths

$$\eqalign{ & {2^2} + {4^2} - 6 \times 2 - 10 \times 4 + \lambda < 0 \cr & \Rightarrow \,\lambda - 32 < 0{\text{ or }}\lambda < 32 \cr & {\text{Solving }}y = 0,\,{x^2} + {y^2} - 6x - 10y + \lambda = 0, \cr & {\text{we get }}{x^2} - 6x + \lambda = 0, \cr & {\text{which must have imaginary roots i}}{\text{.e}}{\text{.,}} \cr & 36 - 4\lambda < 0{\text{ i}}{\text{.e}}{\text{., }}\lambda > 9 \cr & {\text{Solving }}x = 0,\,{x^2} + {y^2} - 6x - 10y + \lambda = 0, \cr & {\text{we get }}{y^2} - 10y + \lambda = 0, \cr & {\text{which must have imaginary roots i}}{\text{.e}}{\text{.,}} \cr & 100 - 4\lambda < 0{\text{ i}}{\text{.e}}{\text{., }}\lambda > 25.{\text{ So 25}} < \lambda < 32. \cr} $$

Here $${r_1} + {r_2} > d.$$   So, the circles cut each other.

Centre of $$S:O\left( { - 3,\,2} \right)$$   centre of given circle $$A\left( {2,\, - 3} \right)$$
$$ \Rightarrow OA = 5\sqrt 2 $$
Also $$AB=5$$   ($$\because AB = r$$   of the given circle)
$$ \Rightarrow $$ Using Pythagoras theorem in $$\Delta OAB$$
$$r = 5\sqrt 3 $$

For the first circle, centre $$ = \left( {0,\,0} \right),$$  radius $${r_1} = 2$$
For the second circle, centre $$ = \left( {3,\,4} \right),$$  radius $${r_2} = \sqrt {{3^2} + {4^2} + 24} = 7$$
$$\therefore \,\left| {{r_1} - {r_2}} \right| = d$$     ( $$=$$ distance between centres ). So the first circle touches the second internally.

The other end is $$\left( {t,\,3 - t} \right)$$
So the equation of the variable circle is
$$\eqalign{ & \left( {x - 1} \right)\left( {x - t} \right) + \left( {y - 1} \right)\left( {y - 3 + t} \right) = 0 \cr & {\text{or }}{x^2} + {y^2} - \left( {1 + t} \right)x - \left( {4 - t} \right)y + 3 = 0 \cr} $$
$$\therefore $$  the centre $$\left( {\alpha ,\,\beta } \right)$$  is given by $$\alpha = \frac{{1 + t}}{2},\,\,\beta = \frac{{4 - t}}{2}$$
$$ \Rightarrow \,\,2\alpha + 2\beta = 5$$

The given circle is $${x^2} + {y^2} - 2x + 4y + 3 = 0.$$      Centre $$\left( {1,\, - 2} \right).$$  Lines through centre $$\left( {1,\, - 2} \right)$$  and parallel to axes are $$x=1$$  and $$y=-2$$

Let the side of square be $$2k.$$
Then sides of square are $$x =1-k$$   and $$x=1+k$$
and $$y=-2-k$$   and $$y=-2+k$$
$$\therefore $$ Co-ordinates of $$P,\,Q,\,R, \,S$$    are $$\left( {1 + k,\, - 2 + k} \right),\,\left( {1 - k,\, - 2 + k} \right),\,\left( {1 - k,\, - 2 - k} \right),\,\left( {1 + k,\, - 2 - k} \right)$$             respectively.
Also $$P\left( {1 + k,\, - 2 + k} \right)$$    lies on circle
$$\eqalign{ & \therefore {\left( {1 + k} \right)^2} + {\left( { - 2 + k} \right)^2} - 2\left( {1 + k} \right) + 4\left( { - 2 + k} \right) + 3 = 0 \cr & \Rightarrow 2{k^2} = 2 \cr & \Rightarrow k = 1\,\,{\text{or}}\,\, - 1 \cr & {\text{If }}k = 1, \cr & P\left( {2,\, - 1} \right),\,Q\left( {0,\, - 1} \right),\,R\left( {0,\, - 3} \right),\,S\left( {2,\, - 3} \right) \cr & {\text{If }}k = - 1, \cr & P\left( {0,\, - 3} \right),\,Q\left( {2,\, - 3} \right),\,R\left( {2,\, - 1} \right),\,S\left( {0,\, - 1} \right) \cr} $$

Two diameters are along $$2x + 3y + 1 = 0$$   and $$3x-y-4=0$$
solving we get centre $$\left( {1,\, - 1} \right)$$
circumference $$ = 2\pi r = 10\pi $$
$$\therefore r = 5$$
Required circle is, $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {5^2}$$
$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 23 = 0$$

$$y = \sqrt {25 - {x^2}} ,\,y = 0$$     bound the semicircle above the $$x$$-axis.
$$\therefore \,\,a + 1 > 0{\text{ and }}{a^2} + {\left( {a + 1} \right)^2} - 25 < 0$$
Solve these and take the common values.

Equation of circle having radius $$r$$ and centre $$\left( {3,\,4} \right)$$  is $$ = {\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} = {r^2}$$
if it is passing through $$\left( {0,\,0} \right)$$
$$\eqalign{ & \therefore \,{\left( {0 - 3} \right)^2} + {\left( {0 - 4} \right)^2} = {r^2} \cr & \Rightarrow {r^2} = 25 \cr} $$
Equation of circle is $${\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} = 25$$
Putting $$y = 0$$
$$\therefore \,x = 6$$   units $$=$$ interception $$x$$-axis intercept on $$y$$ axis (putting $$x = 0$$  ) is $$y = 8$$  units.

$$\eqalign{ & {\text{Here radius}} = \sqrt {{{\left( {\frac{\lambda }{2}} \right)}^2} + {{\left( {\frac{{1 - \lambda }}{2}} \right)}^2} - 5} \leqslant 5 \cr & \Rightarrow 2{\lambda ^2} - 2\lambda - 119 \leqslant 0 \cr & \therefore \,\,\frac{{1 - \sqrt {239} }}{2} \leqslant \lambda \leqslant \frac{{1 + \sqrt {239} }}{2} \cr & \Rightarrow - 7.2 \leqslant \lambda \leqslant 8.2\,\left( {{\text{nearly}}} \right)\,\,\,\,\,\,\,\therefore \,\,\lambda = - 7,\, - 6,\,......,\,8. \cr} $$