Circle MCQ Questions & Answers in Geometry | Maths

We have two circles $${\left( {x - 1} \right)^2} + {\left( {y - 3} \right)^2} = {r^2}$$
Centre (1, 3), radius $$=r$$
and $${x^2} + {y^2} - 8x + 2y + 8 = 0$$
Centre $$\left( {4,\, - 1} \right),$$  radius $$ = \sqrt {16 + 1 - 8} = 3$$

As the two circles intersect each other in two distinct points we should have
$$\eqalign{ & {C_1}{C_2} < {r_1} + {r_2}\,\,\,\,\,{\text{and}}\,\,\,\,\,\,{C_1}{C_2} > \left| {{r_1} - {r_2}} \right| \cr & \Rightarrow {C_1}{C_2} < r + 3\,\,\,\,\,{\text{and}}\,\,\,\,\,\,{C_1}{C_2} < \left| {{r_1} - {r_2}} \right| \cr & \Rightarrow \sqrt {9 + 16} < r + 3\,\,\,\,\,{\text{and}}\,\,\,\,\,\,5 > \left| {r - 3} \right| \cr & \Rightarrow 5 < r + 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \left| {r - 3} \right| < 5 \cr & \Rightarrow r > 2.....({\text{i}})\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow - 5 < r - 3 < 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow - 2 < r < 8.....({\text{ii}}) \cr} $$
Combining (i) and (ii), we get
$$2 < r < 8$$

The given point is an interior point
$$\eqalign{ & \Rightarrow {\left( { - 5 + \frac{r}{{\sqrt 2 }}} \right)^2} + {\left( { - 3 + \frac{r}{{\sqrt 2 }}} \right)^2} - 16 < 0 \cr & \Rightarrow {r^2} - 8\sqrt 2 r + 18 < 0 \cr & \Rightarrow 4\sqrt 2 - \sqrt {14} < r < 4\sqrt 2 + \sqrt {14} \cr} $$
The point is on the major segment $$ \Rightarrow $$ the centre and the point are on the same side of the line $$x + y = 2.$$
$$\eqalign{ & \therefore \,\, - 5 + \frac{r}{{\sqrt 2 }} - 3 + \frac{r}{{\sqrt 2 }} - 2 < 0 \cr & \Rightarrow \,r < 5\sqrt 2 {\text{ So, }}4\sqrt 2 - \sqrt {14} < r < 5\sqrt 2 \cr} $$

Let any tangent to circle $${x^2} + {y^2} = 1$$   is $$x\,\cos \,\theta + y\,\sin \,\theta = 1$$
Since, $$P$$ and $$Q$$ are the point of intersection on the co-ordinate axes.
Then $$P \equiv \left( {\frac{1}{{\cos \,\theta }},\,0} \right)\,\,\& \,\,Q \equiv \left( {0,\,\frac{1}{{\sin \,\theta }}} \right)$$
$$\therefore $$ mid-point of $$PQ$$  be
$$\eqalign{ & M \equiv \left( {\frac{1}{{2\,\cos \,\theta }},\,\,\frac{1}{{2\,\sin \,\theta }}} \right) \equiv \left( {h,\,k} \right) \cr & \Rightarrow \cos \,\theta = \frac{1}{{2h}}.....(1) \cr & \Rightarrow \sin \,\theta = \frac{1}{{2k}}.....(2) \cr} $$
Now squaring and adding equation (1) and (2)
$$\eqalign{ & \frac{1}{{{h^2}}} + \frac{1}{{{k^2}}} = 4 \cr & \Rightarrow {h^2} + {k^2} = 4{h^2}{k^2} \cr} $$
$$\therefore $$ locus of $$M$$ is : $${x^2} + {y^2} - 4{x^2}{y^2} = 0$$

The given circle is $${x^2} + {y^2} - 6x - 6y + 14 = 0,$$     centre (3, 3), radius $$= 2$$
Let $$\left( {h,\,k} \right)$$  be the centre of touching circle. Then radius of touching circle $$=h$$ [as it touches $$y$$-axis also]
$$\therefore $$ Distance between centres of two circles $$=$$ sum of the radii of two circles
$$\eqalign{ & \Rightarrow \sqrt {{{\left( {h - 3} \right)}^2} + {{\left( {k - 3} \right)}^2}} = 2 + h \cr & \Rightarrow {\left( {h - 3} \right)^2} + {\left( {k - 3} \right)^2} = {\left( {2 + h} \right)^2} \cr & \Rightarrow {k^2} - 10h - 6k + 14 = 0 \cr} $$
$$\therefore $$ locus of $$\left( {h,\,k} \right)$$  is $${y^2} - 10x - 6y + 14 = 0$$

$${a^2} + {4^2} + 10a > 0,\,{a^2} + {4^2} - 12a + 20 > 0,$$         i.e., the inequations in $$a$$ are $${a^2} + 10a + 16 > 0$$    and $${a^2} - 12a + 36 > 0.$$    Solve these and take the common values.

Here, the centre $$ = \left( { - 2,\,3} \right)$$   and the radius $$ = \sqrt {{{\left( { - 2} \right)}^2} + {3^2} - 9{{\sin }^2}\alpha - 13{{\cos }^2}\alpha } = 2\sin \,\alpha $$
$$\eqalign{ & \therefore \,\,\sin \,\alpha = \frac{{2\sin \,\alpha }}{{\sqrt {{{\left( {{x_1} + 2} \right)}^2} + {{\left( {{y_1} - 3} \right)}^2}} }} \cr & \Rightarrow \,\,{\left( {{x_1} + 2} \right)^2} + {\left( {{y_1} - 3} \right)^2} = 4\, \cr} $$

Since the $$\left( {\left[ {P + 1} \right],\,\left[ P \right]} \right)$$    lies inside the circle
$$\eqalign{ & {x^2} + {y^2} - 2x - 15 = 0\,\,\left[ {{\text{But }}\left[ {x + n} \right] = \left[ x \right] + n,\,n\, \in \,N} \right] \cr & \therefore \,{\left[ {P + 1} \right]^2} + {\left[ P \right]^2} - 2\left[ {P + 1} \right] - 15 < 0 \cr & \Rightarrow {\left( {\left[ P \right] + 1} \right)^2} + {\left[ P \right]^2} - 2\left( {\left[ P \right] + 1} \right) - 15 < 0 \cr & \Rightarrow 2{\left[ P \right]^2} - 16 < 0 \cr & \Rightarrow {\left[ P \right]^2} < 8......\left( 1 \right) \cr} $$
From the second circle
$$\eqalign{ & {\left( {\left[ P \right] + 1} \right)^2} + {\left[ P \right]^2} - 2\left( {\left[ P \right] + 1} \right) - 7 > 0 \cr & \Rightarrow 2{\left[ P \right]^2} - 8 > 0 \cr & \Rightarrow {\left[ P \right]^2} > 4......\left( 2 \right) \cr} $$
From $$\left( 1 \right)\& \left( 2 \right),\,\,4 < {\left[ P \right]^2} < 8,$$     which is not possible.
$$\therefore $$  for no values of $$'P'$$ the point will be within the region.

$$2gg' + 2ff' = 0 + 0 = 25 - 25 = c + c'$$
$$\therefore $$  the circles cut orthogonally.

The vertices are $$A\left( {r,\,0} \right),\,B\left( {r\cos \,{{60}^ \circ },\,r\sin \,{{60}^ \circ }} \right),\,C\left( { - r\cos \,{{60}^ \circ },\,r\sin \,{{60}^ \circ }} \right),\,D\left( { - r,\,0} \right),\,E\left( { - r\cos \,{{60}^ \circ },\, - r\sin \,{{60}^ \circ }} \right),\,F\left( {r\cos \,{{60}^ \circ },\, - r\sin \,{{60}^ \circ }} \right).$$
If $$P = \left( {x,\,y} \right)$$   then $$\sum {P{A^2} = 6{a^2}} $$    implies
$$\eqalign{ & \left\{ {{{\left( {x - r} \right)}^2} + {y^2}} \right\} + \left\{ {{{\left( {x - \frac{r}{2}} \right)}^2} + {{\left( {y - \frac{{r\sqrt 3 }}{2}} \right)}^2}} \right\} + ..... + \left\{ {{{\left( {x - \frac{r}{2}} \right)}^2} + {{\left( {x + \frac{{r\sqrt 3 }}{2}} \right)}^2}} \right\} = 6{a^2} \cr & {\text{or }}2\left( {{x^2} + {y^2} + {r^2}} \right) + 2\left( {{x^2} + {y^2} + \frac{{{r^2}}}{4} + \frac{{3{r^2}}}{4}} \right) + 2\left( {{x^2} + {y^2} + \frac{{{r^2}}}{4} + \frac{{3{r^2}}}{4}} \right) = 6{a^2} \cr & {\text{or }}{x^2} + {y^2} + {r^2} = {a^2} \cr & {\text{or }}{x^2} + {y^2} = {\left( {\sqrt {{a^2} - {r^2}} } \right)^2} \cr} $$

$$\eqalign{ & \left| {{r_1} - {r_2}} \right| < {C_1}{C_2}{\text{ for intersection}} \cr & \Rightarrow r - 3 < 5 \Rightarrow r < 8.....\left( 1 \right) \cr & {\text{and }}{r_1} + {r_2} > {C_1}{C_2},\,\,\,r + 3 > 5 \Rightarrow r > 2.....\left( 2 \right) \cr & {\text{From }}\left( 1 \right){\text{ and }}\left( 2 \right),{\text{ we get}} \cr & 2 < r < 8 \cr} $$