Circle MCQ Questions & Answers in Geometry | Maths

For the first circle, centre $$ = \left( {3,\,7} \right),$$   radius $${r_1} = \sqrt {{3^2} + {7^2} - 48} = \sqrt {10} $$
For the second circle, centre $$ = \left( {3,\,0} \right),$$   radius $${r_2} = 3$$
So, $${r_1} + {r_2} < d$$   ( $$=$$ the distance between centres ).
$$\therefore $$  circles do not cut and hence the number of common tangents $$=4.$$

For the given circle,
centre : $$\left( {4,\,4} \right)$$
radius $$= 6$$
$$\eqalign{ & 6 + k = \sqrt {{{\left( {h - 4} \right)}^2} + {{\left( {k - 4} \right)}^2}} \cr & {\left( {h - 4} \right)^2} = 20k + 20 \cr} $$
locus of $$\left( {h,\,k} \right)$$  is
$${\left( {x - 4} \right)^2} = 20\left( {y + 1} \right),$$
which is a parabola.

Circles are $${\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 2$$     and $$x\left( {x - 1} \right) + y\left( {y - 1} \right) = 0.$$     Here $$\left| {{r_1} - {r_2}} \right| = d.$$

As centre of one circle is (0, 0) and other circle passes through (0, 0), therefore
$$\eqalign{ & {\text{Also }}{C_1}\left( {\frac{a}{2},\,0} \right)\,\,\,\,{C_2}\left( {0,\,0} \right) \cr & {r_1} = \frac{a}{2}\,\,\,\,{r_2} = C \cr & {C_1}{C_2} = {r_1} - {r_2} = \frac{a}{2} \cr & \Rightarrow C - \frac{a}{2} = \frac{a}{2} \cr & \Rightarrow C = a \cr} $$
If the two circles touch each other, then they must touch each other internally.

Equation of circle $$C \equiv {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 1$$
Radius of $$T = \left| y \right|$$
$$T$$ touches $$C$$ externally
therefore,
Distance between the centres $$=$$ sum of their radii
$$\eqalign{ & \Rightarrow \sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {y - 1} \right)}^2}} = 1 + \left| y \right| \cr & \Rightarrow {\left( {0 - 1} \right)^2} + {\left( {y - 1} \right)^2} = {\left( {1 + \left| y \right|} \right)^2} \cr & \Rightarrow 1 + {y^2} + 1 - 2y = 1 + {y^2} + 2\left| y \right| \cr & \Rightarrow 2\left| y \right| = 1 - 2y \cr & {\text{If }}y > 0{\text{ then 2}}y = 1 - 2y \Rightarrow y = \frac{1}{4} \cr & {\text{If }}y < 0{\text{ then }} - {\text{2}}y = 1 - 2y \Rightarrow 0 = 1\left( {{\text{not possible}}} \right) \cr & \therefore \,\,\,\,y = \frac{1}{4} \cr} $$

Let the equation of the circle be
$${x^2} + {y^2} + 2gx + 2fy + c = 0$$
As this circle passes through $$\left( {0,\,0} \right)$$  and $$\left( {1,\,0} \right),$$  we get $$c=0,\,\,1+2g=0$$
$$ \Rightarrow g = - \frac{1}{2}$$
According to the question this circle touches the given circle $${x^2} + {y^2} = 9$$
$$ \Rightarrow 2 \times $$   radius of required circle $$=$$ radius of given circle
$$\eqalign{ & \Rightarrow 2\sqrt {{g^2} + {f^2}} = 3 \cr & \Rightarrow {g^2} + {f^2} = \frac{9}{4} \cr & \Rightarrow \frac{1}{4} + {f^2} = \frac{9}{4} \cr & \Rightarrow {f^2} = 2\,\,\,\,\, \cr & \Rightarrow f = \pm \sqrt 2 \cr} $$
$$\therefore $$ The centre is $$\left( {\frac{1}{2},\,\sqrt 2 } \right),\,\,\left( {\frac{1}{2},\, - \sqrt 2 } \right)$$

Let the centre of the circle be $$\left( {h,\,2} \right)$$
$$\therefore $$ Equation of circle is
$${\left( {x - h} \right)^2} + {\left( {y - 2} \right)^2} = 25.....(1)$$
Differentiating with respect to $$x,$$  we get
$$\eqalign{ & 2\left( {x - h} \right) + 2\left( {y - 2} \right)\frac{{dy}}{{dx}} = 0 \cr & \Rightarrow x - h = - \left( {y - 2} \right)\frac{{dy}}{{dx}} \cr} $$
Substituting in equation (1), we get
$$\eqalign{ & {\left( {y - 2} \right)^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25 \cr & \Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2} \cr} $$

For any point $$P\left( {x,\,y} \right)$$   in the given circle,

we should have
$$\eqalign{ & OA \leqslant OP \leqslant OB \Rightarrow \left( {5 - 3} \right) \leqslant \sqrt {{x^2} + {y^2}} \leqslant 5 + 3 \cr & \Rightarrow 4 \leqslant {x^2} + {y^2} \leqslant 64 \cr} $$

Circle $${x^2} + {y^2} - 4x - 8y - 5 = 0$$
Centre $$ = \left( {2,\,4} \right),$$   Radius $$ = \sqrt {4 + 16 + 5} = 5$$
If circle is intersecting line $$3x-4y=m,$$    at two distinct points.
$$ \Rightarrow $$ length of perpendicular from centre to the line < radius
$$\eqalign{ & \Rightarrow \frac{{\left| {6 - 16 - m} \right|}}{5} < 5 \cr & \Rightarrow \left| {10 + m} \right| < 25 \cr & \Rightarrow - 25 < m + 10 < 25 \cr & \Rightarrow - 35 < m < 15 \cr} $$

$$\eqalign{ & {\text{Let }}\angle RPS = \theta \cr & \,\,\,\,\,\,\,\,\angle XPQ = {90^ \circ } - \theta \cr} $$

$$\eqalign{ & \therefore \angle PQX = \theta \,\,\,\,\,\,\,\,\left( {\because \,\angle PXQ = {{90}^ \circ }} \right) \cr & \therefore \Delta PRS \sim \Delta QPR\,\,\,\,\,\,\left( {{\text{AA similarity}}} \right) \cr & \therefore \frac{{PR}}{{QP}} = \frac{{RS}}{{PR}} \Rightarrow P{R^2} = PQ.RS \cr & \Rightarrow PR = \sqrt {PQ.RS} \Rightarrow 2r = \sqrt {PQ.RS} \cr} $$