Circle MCQ Questions & Answers in Geometry | Maths

Solving $$y =x$$  and the circle
$${x^2} + {y^2} - 2x = 0,$$    we get
$$x=0,\,\,y=0$$   and $$x=1,\,\,y=1$$
$$\therefore $$ Extremities of diameter of the required circle are $$\left( {0,\,0} \right)$$  and $$\left( {1,\,1} \right).$$
Hence, the equation of circle is
$$\eqalign{ & \left( {x - 0} \right)\left( {x - 1} \right) + \left( {y - 0} \right)\left( {y - 1} \right) = 0 \cr & \Rightarrow {x^2} + {y^2} - x - y = 0 \cr} $$

For $${x^2} + {y^2} = 9,$$   the centre $$ = \left( {0,\,0} \right)$$  and the radius $$= 3.$$
For $${x^2} + {y^2} - 8x - 6y + {n^2} = 0,$$      the centre $$ = \left( {4,\,3} \right)$$  and the radius $$ = \sqrt {{4^2} + {3^2} - {n^2}} $$
$$\therefore \,{4^2} + {3^2} - {n^2} > 0{\text{ or }}{n^2} < {5^2}{\text{ or }} - 5 < n < 5$$
Circles should cut to have exactly two common tangents. So, $${r_1} + {r_2} > d$$
$$\eqalign{ & \therefore \,\,3 + \sqrt {25 - {n^2}} \, > \,\sqrt {{4^2} + {3^2}} {\text{ or }}\sqrt {25 - {n^2}} > 2{\text{ or }}25 - {n^2} > 4 \cr & \therefore \,\,{n^2} < 21{\text{ or }} - \sqrt {21} < n < \sqrt {21} \cr} $$
Therefore, common values of $$n$$ should satisfy $$ - \sqrt {21} < n < \sqrt {21} .$$    But $$n\, \in \,Z.$$
So, $$n = - 4,\, - 3,\,.....,\, 4.$$

Radius of the circle $$=$$ perpendicular distance of $$\left( {2,\,3} \right)$$  from $$x + y = 1$$   is $$\frac{4}{{\sqrt 2 }} = 2\sqrt 2 $$
$$\therefore $$  The required equation will be
$$\eqalign{ & {\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} = 8 \cr & \Rightarrow {x^2} + {y^2} - 4x - 6y + 5 = 0 \cr} $$

Any circle passing through the points of intersection of the given circles is $${x^2} + {y^2} - 2x + y + \lambda \left( {{x^2} + {y^2} - 1} \right) = 0.$$        If $$\left( {\alpha ,\,\beta } \right)$$  be its centre then $$\alpha = \frac{1}{{1 + \lambda }},\,\,\beta = \frac{{ - \frac{1}{2}}}{{1 + \lambda }}\,\,\, \Rightarrow \alpha + 2\beta = 0.$$

The centre of $${C_1} = \left( {2,\,1} \right)$$   and the radius $$ = \sqrt {{2^2} + {1^2} + 11} = 4$$
If $$\left( {\alpha ,\,\beta } \right)$$  be the centre of $${C_2},\,\,\sqrt {{{\left( {\alpha - 2} \right)}^2} + {{\left( {\beta - 1} \right)}^2}} = 4 + 1.$$

$$\eqalign{ & {s_1} = {x^2} + {y^2} + 2ax + cy + a = 0 \cr & {s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0 \cr} $$
Equation of common chord of circles $${s_1}$$ and $${s_2}$$ is given by $${s_1} - {s_2} = 0$$
$$ \Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0$$
Given that $$5x + by- a =0$$    passes through $$P$$ and $$Q$$
$$\therefore $$ The two equations should represent the same line
$$\eqalign{ & \Rightarrow \frac{a}{1} = \frac{{c - d}}{b} = \frac{{a + 1}}{{ - a}} \cr & \Rightarrow a + 1 = - {a^2} \cr & \Rightarrow {a^2} + a + 1 = 0 \cr} $$
No real value of $$a$$

The equation of the circle $$S$$ is
$${x^2} + {\left( {y - \sqrt 2 } \right)^2} = {r^2}......\left( 1 \right)$$
Let the coordinates of any point on this circle be $$\left( {h,\,k} \right),$$  then
$$\eqalign{ & {h^2} + {\left( {k - \sqrt 2 } \right)^2} = {r^2} \cr & \Rightarrow k = \sqrt 2 \pm \sqrt {{r^2} - {h^2}} ......\left( 2 \right) \cr} $$
Since the above value of $$k$$ contains a constant irrational number $$\sqrt 2 ,$$  therefore, the only possible rational value of $$k$$ is $$0.$$ Hence,
$$\eqalign{ & \sqrt 2 \pm \sqrt {{r^2} - {h^2}} = 0 \cr & \Rightarrow {r^2} - {h^2} = 2 \cr & \Rightarrow h = \pm \sqrt {{r^2} - 2} \cr} $$
Thus, we have following cases
$$\left( {\bf{i}} \right)$$   If $${r^2} - 2$$  is a perfect square, there will be two rational points, viz., $$\left( {\sqrt {{r^2} - 2} ,\,0} \right)$$   and $$\left( { - \sqrt {{r^2} - 2} ,\,0} \right)$$   on $$S.$$
$$\left( {{\bf{ii}}} \right)$$   If $${r^2} - 2$$  is not a perfect square, there will be no rational point on $$S.$$
$$\therefore $$  there can be at most two rational points on $$S.$$

Let $$A \equiv \left( {2,\,0} \right)$$
Given lines are $$3x + 5y = 1......\left( 1 \right)$$
and $$\left( {2 + c} \right)x + 5{c^2}y = 1......\left( 2 \right)$$
Multiplying equation $$\left( 1 \right)$$ by $${c^2}$$ and subtracting $$\left( 2 \right)$$ form it, we get
$$\eqalign{ & \left( {3{c^2} - c - 2} \right)x = {c^2} - 1 \cr & {\text{or }}x = \frac{{{c^2} - 1}}{{3{c^2} - c - 2}} \cr} $$
Now, $$\mathop {\lim }\limits_{c \to 1} \,x = \mathop {\lim }\limits_{c \to 1} \frac{{\left( {c - 1} \right)\left( {c + 1} \right)}}{{\left( {c - 1} \right)\left( {3c + 2} \right)}} = \mathop {\lim }\limits_{c \to 1} \frac{{c + 1}}{{3c + 2}} = \frac{2}{5}$$
$$\therefore \,x$$  coordinate of centre $$ = \frac{2}{5}$$
From equation $$\left( 1 \right),$$ when $$x = \frac{2}{5},\,y = - \frac{1}{{25}}$$
Hence, the centre of the circle is $$\left( {\frac{2}{5},\, - \frac{1}{{25}}} \right)$$
Also, the circle passes through the point $$A\left( {2,\,0} \right)$$
$$\therefore $$  radius of the circle $$ = \sqrt {{{\left( {2 - \frac{2}{5}} \right)}^2} + {{\left( {0 + \frac{1}{{25}}} \right)}^2}} $$
Thus, equation of the required circle is
$$\eqalign{ & = {\left( {x - \frac{2}{5}} \right)^2} + {\left( {y + \frac{1}{{25}}} \right)^2} = \frac{{64}}{{25}} + \frac{1}{{625}} \cr & {\text{or }}25\left( {{x^2} + {y^2}} \right) - 20x + 2y - 60 = 0 \cr} $$

Circle $${\left( {x - 3} \right)^2} + {\left( {y + 2} \right)^2} = 25,$$     with centre $$C\left( {3,\, - 2} \right)$$   and radius $$5$$ is intersected by a line $$y= mx + 1$$   at $$P\,\& \,Q$$  such that mid point $$R$$ of $$PQ$$  has its $$x$$-co-ordinate as $$ - \frac{3}{5}.$$
Let $$R\left( - {\frac{3}{5},\,\frac{{ - 3m}}{5} + 1} \right)$$

$$\eqalign{ & {\text{Then }}CR \bot PQ \Rightarrow \frac{{\frac{{ - 3m}}{5} + 1 + 2}}{{ - \frac{3}{5} - 3}} \times m = - 1 \cr & \Rightarrow \frac{{\left( { - 3m + 15} \right)m}}{{ - 18}} = - 1 \cr & \Rightarrow {m^2} - 5m + 6 = 0 \cr & \Rightarrow m = 2,\,3 \cr} $$
Alternately :-
The $$x$$-co-ordinates of intersection points $$P$$ and $$Q$$ of circle and line can be given by solving their equations for $$x.$$  Putting $$y= mx + 1$$   in $${\left( {x - 3} \right)^2} + {\left( {y + 2} \right)^2} = 25,$$     we get
$$\eqalign{ & {\left( {x - 3} \right)^2} + {\left( {mx + 3} \right)^2} = 25 \cr & \Rightarrow \left( {{m^2} + 1} \right){x^2} + 6\left( {m - 1} \right)x - 16 = 0 \cr & \Rightarrow {x_1} + {x_2} = \frac{{ - 6\left( {m - 1} \right)}}{{{m^2} + 1}} \cr} $$
where $${x_1}$$ and $${x_2}$$ are x-coordinates of $$P$$ and $$Q$$ respectively.
As per question, x-coordinate of mid point of $$PQ$$  is $$ - \frac{3}{5}$$
$$\eqalign{ & \therefore \,\,\frac{{{x_1} + {x_2}}}{2} = - \frac{3}{5} \cr & \Rightarrow \frac{{ - 3\left( {m - 1} \right)}}{{{m^2} + 1}} = - \frac{3}{5} \cr & \Rightarrow {m^2} - 5m + 6 = 0 \cr & \Rightarrow m = 2,\,3 \cr} $$

Let the second circle be $${x^2} + {y^2} + 2gx + 2fy = 0.$$     The common chord has the equation $$2\left( {g - 3} \right)x + 2\left( {f - 4} \right)y + 7 = 0.$$
But $$y = x$$  touches the circle.
Hence, $${x^2} + {x^2} + 2gx + 2fx = 0$$      has equal roots, i.e., $$f + g = 0.$$
$$\therefore $$  the equation of the common chord is $$2\left( {g - 3} \right)x + 2\left( { - g - 4} \right)y + 7 = 0$$
or $$\left( { - 6x - 8y + 7} \right) + g\left( {2x - 2y} \right) = 0,$$       which passes through the point of intersection of $$ - 6x - 8y + 7 = 0$$     and $$2x - 2y = 0.$$