Circle MCQ Questions & Answers in Geometry | Maths

The centre of the given circle is $$\left( { - 8,\,12} \right)$$  and radius is $$5.$$

The image of the circle will have the same radius, i.e. the radius of the required circle is $$5$$. The centre $$D$$ of the required circle is the image of the centre $$C$$ of the given circle in the line mirror. If $$D$$ be $$\left( {\alpha ,\,\beta } \right)$$  then
$$\eqalign{ & \frac{{\alpha + 8}}{4} = \frac{{\beta - 12}}{7} = - 2\left[ {\frac{{4 \times - 8 + 7 \times 12 + 13}}{{{4^2} + {7^2}}}} \right]\,\,\,\,\left[ {{\text{See straight line}}} \right] \cr & {\text{or, }}\frac{{\alpha + 8}}{4} = \frac{{\beta - 12}}{7} = \frac{{ - 2 \times 65}}{{65}} = - 2 \cr & \therefore \,\alpha = - 16,\,\,\beta = - 2 \cr} $$
$$\therefore $$  Required circle is $${\left( {x + 16} \right)^2} + {\left( {y + 2} \right)^2} = {5^2}$$

Point of intersection of $$3x-4y-7=0$$    and $$2x-3y-5=0$$    is (1, $$-$$ 1) which is the centre of the circle and radius $$=7$$
$$\therefore $$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$
$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$

Equation of circle with centre $$\left( {0,\,3} \right)$$  and radius $$2$$ is $${x^2} + {\left( {y - 3} \right)^2} = 4$$
Let locus of the variable circle is $$\left( {\alpha ,\,\beta } \right)$$
$$\because $$ It touches $$x$$-axis.
$$\therefore $$ It's equation is $${\left( {x - \alpha } \right)^2} + {\left( {y + \beta } \right)^2} = {\beta ^2}$$

Circle touch externally $$ \Rightarrow {c_1}{c_2} = {r_1} + {r_2}$$
$$\eqalign{ & \therefore \sqrt {{\alpha ^2} + {{\left( {\beta - 3} \right)}^2}} = 2 + \beta \cr & {\alpha ^2} + {\left( {\beta - 3} \right)^2} = {\beta ^2} + 4 + 4\beta \cr & \Rightarrow {\alpha ^2} = 10\left( {\beta - \frac{1}{2}} \right) \cr} $$
$$\therefore $$ Locus is $${x^2} = 10\left( {y - \frac{1}{2}} \right)$$    which is a parabola.

Given combined equation is
$$\eqalign{ & \left( {x + y - 4} \right)\left( {xy - 2x - y + 2} \right) = 0 \cr & \Rightarrow \left( {x + y - 4} \right)\left( {x - 1} \right)\left( {y - 2} \right) = 0 \cr} $$
So, the equation of sides of the triangle are $$x = 1,\,y = 2,\,x + y = 4$$
Point of intersection of any two sides of a triangle will give the vertices of triangle.
So, the vertices of the triangle are $$A\left( {1,\,3} \right),\,B\left( {1,\,2} \right),\,C\left( {2,\,2} \right)$$
Slope of $$AC = \frac{{2 - 3}}{{2 - 1}} = - 1$$
So, slope of its perpendicular bisector is 1.
Mid-point of $$AC$$  is $$\left( {\frac{3}{2},\,\frac{5}{2}} \right)$$
So, equation of perpendicular bisector of $$AC$$  is
$$\eqalign{ & y - \frac{5}{2} = \left( {x - \frac{3}{2}} \right) \cr & \Rightarrow 2y - 2x = 2 \cr & \Rightarrow y - x = 1......\left( 1 \right) \cr} $$
Clearly, $$AB$$  makes an angle of $${90^ \circ }$$  with $$x$$-axis.
So, slope of perpendicular bisector of $$AB$$  is 0.
Mid-point of $$AB$$  is $$\left( {1,\,\frac{5}{2}} \right)$$
So, equation of perpendicular bisector of $$AB$$  is
$$y - \frac{5}{2} = 0......\left( 2 \right)$$
Solving equation (1) and (2), we get
$$x = \frac{3}{2},\,y = \frac{5}{2}$$
So, the coordinates of circumcenter is $$O\left( {\frac{3}{2},\,\frac{5}{2}} \right)$$
Radius $$ = OA = \frac{1}{{\sqrt 2 }}$$
Equation of circumcircle is
$$\eqalign{ & {\left( {x - \frac{3}{2}} \right)^2} + {\left( {y - \frac{5}{2}} \right)^2} = \frac{1}{2} \cr & \Rightarrow 4{x^2} + 9 - 12x + 4{y^2} + 25 - 20y = 2 \cr & \Rightarrow {x^2} + {y^2} - 3x - 5y + 8 = 0 \cr} $$

$$\eqalign{ & \left( {{x_{\text{i}}},\,{y_{\text{i}}}} \right),\,{\text{i}} = 1,\,2,\,3,\,4{\text{ lies on}} \cr & xy = {c^2} \Rightarrow {y_{\text{i}}} = \frac{{{c^2}}}{{{x_{\text{i}}}}} \cr & {\text{Now the point }}\left( {{x_{\text{i}}},\,{y_{\text{i}}}} \right){\text{ lies on}} \cr & {x^2} + {y^2} = {a^2} \cr & \Rightarrow x_{\text{i}}^2 + \frac{{{c^4}}}{{{x_{\text{i}}}}} = {a^2} \cr & \Rightarrow x_{\text{i}}^4 - {a^2}x_{\text{i}}^2 + {c^4} = 0 \cr & {\text{Its roots are }}{x_1},\,{x_2},\,{x_3},\,{x_4} \cr & \therefore \,{x_1} + {x_2} + {x_3} + {x_4} = 0 \cr & \Rightarrow {x_1}{x_2} + {x_1}{x_3} + {x_1}{x_4} + {x_2}{x_3} + {x_2}{x_4} + {x_3}{x_4} = {a^2} \cr & \Rightarrow {x_1}{x_2}{x_3} + {x_1}{x_2}{x_4} + {x_1}{x_3}{x_4} + {x_2}{x_3}{x_4} = 0 \cr & \Rightarrow {x_1}{x_2}{x_3}{x_4} = {c^4}{\text{ Clearly }}\left( {\bf{C}} \right){\text{ is not correct}} \cr & {\text{Now, }}{y_1} + {y_2} + {y_3} + {y_4} = \frac{{{c^2}}}{{{x_1}}}.\frac{{{c^2}}}{{{x_2}}}.\frac{{{c^2}}}{{{x_3}}}.\frac{{{c^2}}}{{{x_4}}} = {c^4} \cr & {\text{and }}{y_1} + {y_2} + {y_3} + {y_4} = \frac{{{c^2}\left( {\sum {{x_1}{x_2}{x_3}} } \right)}}{{{x_1}{x_2}{x_3}{x_4}}} = 0 \cr} $$

Equations of the circles are
$$\eqalign{ & {x^2} + {y^2} = 4......\left( 1 \right) \cr & {\text{and }}{x^2} + {y^2} = 2x + 2y......\left( 2 \right) \cr} $$
Centre of $$\left( 1 \right)$$ is $${C_1} \equiv \left( {0,\,0} \right)\,;$$   Radius of $$\left( 1 \right) = {r_1} = 2\,;$$
Centre of $$\left( 2 \right)$$ is $${C_2} \equiv \left( {1,\,1} \right)\,;$$   Radius of $$\left( 2 \right) = {r_2} = \sqrt 2 $$
$$d = $$  distance between centers $$ = {C_1}{C_2} = \sqrt {1 + 1} = \sqrt 2 $$
If $$\theta $$ is the angle of intersection of two circles, then
$$\eqalign{ & \cos \,\theta = \frac{{r_1^2 + r_2^2 - {d^2}}}{{2{r_1}{r_2}}} = \frac{{{{\left( 2 \right)}^2} + {{\left( {\sqrt 2 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}{{2.2\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \cr & \therefore \,\theta = \frac{\pi }{4} \cr} $$

In the beginning, coordinates of the centre $$ = \left( {2,\,2} \right),$$  radius $$= 2.$$  After one complete revolution on the $$x$$-axis in the positive direction, the $$x$$-coordinate increases by $$2\pi r,$$  i.e., $$4\pi ,$$  whereas the $$y$$-coordinate remains the same.

The tangent at the origin is $$y - x = 0.\,B$$    is the image of $$A\left( {1,\,4} \right)$$  by this line.

Here, the centre $$ = \left( {\frac{p}{2},\,\frac{q}{2}} \right)$$
So, $$\left( {p,\,q} \right)$$  and $$\left( {0,\,0} \right)$$  are the ends of a diameter. As the two chords are bisected by the line $$y=0,$$  the chords will cut the circle at the points $$\left( {{x_1},\, - q} \right)$$  and $$\left( {{x_2},\, - q} \right)$$  where $${x_1},\,{x_2}$$  are real.
Clearly, the line joining these points is $$y=-q.$$  Solving $$y=-q$$   and $${x^2} + {y^2} = px + qy,$$     we get $${x^2} - px + 2{q^2} = 0,$$    whose roots are $${x_1},\,{x_2}.$$
For real distinct roots $$D > 0,\,\,{\text{i}}{\text{.e}}{\text{.,}}\,{p^2} - 8{q^2} > 0.$$

Centre of the circle
$${x^2} + {y^2} + 4x - 6y + 9\,{\sin ^2}\alpha + 13\,{\cos ^2}\alpha = 0$$
is $$C\left( { - 2,\,3} \right)$$   and its radius is
$$\eqalign{ & \sqrt {{2^2} + {{\left( { - 3} \right)}^2} - 9\,{{\sin }^2}\alpha + 13\,{{\cos }^2}\alpha } \cr & = \sqrt {4 + 9 - 9\,{{\sin }^2}\alpha + 13\,{{\cos }^2}\alpha } \cr & = 2\,\sin \,\alpha \cr} $$

Let $$P\left( {h,\,k} \right)$$   be any point on the locus. The $$\angle APC = \alpha $$
Also $$\angle PAC = \frac{\pi }{2}$$
That is, triangle $$APC$$  is a right triangle.
Thus, $$\sin \,\alpha = \frac{{AC}}{{PC}} = \frac{{2\,\sin \,\alpha }}{{\sqrt {{{\left( {h + 2} \right)}^2} + {{\left( {k - 3} \right)}^2}} }}$$
$$\eqalign{ & \Rightarrow \sqrt {{{\left( {h + 2} \right)}^2} + {{\left( {k - 3} \right)}^2}} = 2 \cr & \Rightarrow {\left( {h + 2} \right)^2} + {\left( {k - 3} \right)^2} = 4 \cr & {\text{or}}\,\,{h^2} + {k^2} + 4h - 6k + 9 = 0 \cr} $$
Thus required equation of the locus is $${x^2} + {y^2} + 4x - 6y + 9 = 0$$