Circle MCQ Questions & Answers in Geometry | Maths

$$\left( {h,\,k} \right)$$  is a point in the interior of the circle $${x^2} + {y^2} = {a^2}.$$   So, $$hx + ky = {a^2}$$   neither can be a real tangent nor a chord of contact of tangents.

Let centre of the circle be $$\left( {h,\,2} \right)$$  then radius $$ = \left| h \right|$$
$$\therefore $$ Equation of circle becomes $${\left( {x - h} \right)^2} + {\left( {y - 2} \right)^2} = {h^2}$$
As it passes through ($$-$$1, 0)

$$ \Rightarrow {\left( { - 1 - h} \right)^2} + 4 = {h^2} \Rightarrow h = \frac{{ - 5}}{2}$$
$$\therefore $$ Centre $$\left( {\frac{{ - 5}}{2},\,2} \right)$$   and $$r = \frac{5}{2}$$
Distance of centre from ($$-$$ 4, 0) is $$\frac{5}{2}$$
$$\therefore $$ It lies on the circle.

Any point $$P$$ on line $$4x-5y =20$$   is $$\left( {\alpha ,\,\frac{{4\alpha - 20}}{5}} \right).$$
Equation of chord of contact $$AB$$  to the circle $${x^2} + {y^2} = 9$$

drawn from point $$P\left( {\alpha ,\,\frac{{4\alpha - 20}}{5}} \right)$$    is
$$x.\alpha + y.\left( {\frac{{4\alpha - 20}}{5}} \right) = 9.....(1)$$
Also the equation of chord $$AB$$  whose mid point is $$\left( {h,\,k} \right)$$  is
$$hx + ky = {h^2} + {k^2}.....(2)$$
$$\because $$ Equations (1) and (2) represent the same line, therefore
$$\eqalign{ & \frac{h}{\alpha } = \frac{k}{{\frac{{4\alpha - 20}}{5}}} = \frac{{{h^2} + {k^2}}}{9} \cr & \Rightarrow 5k\alpha = 4h\alpha - 20h{\text{ and }}9h = \alpha \left( {{h^2} + {k^2}} \right) \cr & \Rightarrow \alpha = \frac{{20h}}{{4h - 5k}}{\text{ and }}\alpha = \frac{{9h}}{{{h^2} + {k^2}}} \cr & \Rightarrow \frac{{20h}}{{4h - 5k}} = \frac{{9h}}{{{h^2} + {k^2}}} \cr & \Rightarrow 20\left( {{h^2} + {k^2}} \right) = 9\left( {4h - 5k} \right) \cr} $$
$$\therefore $$ Locus of $$\left( {h,\,k} \right)$$
$$20\left( {{x^2} + {y^2}} \right) - 36x + 45y = 0$$

Draw the figure. Clearly, centre $$ = \left( {0,\,\sqrt 3 } \right)$$   and radius $$=2.$$

When two circles $$A$$ and $$B$$ of equal radii pass through the centers of each other, the angle made by arc of $$B$$ at the centre of $$B$$ is $${90^ \circ }.$$
So, length of small arc of $$B = \frac{{2\pi r\,{{90}^ \circ }}}{{{{360}^ \circ }}} = \frac{{\pi r}}{2}$$
Hence, circumference of $$A$$ cut off by the circle $$B = 2\pi r - \frac{{\pi r}}{2} = \frac{{3\pi r}}{2}$$
$$\therefore $$  Required ratio $$ = \frac{{\frac{{\pi r}}{2}}}{{\frac{{3\pi r}}{2}}} = \frac{1}{3}$$

Let the variable circle is
$${x^2} + {y^2} + 2gx + 2fy + c = 0.....(1)$$
It passes through $$\left( {a,\,b} \right)$$
$$\therefore {a^2} + {b^2} + 2ga + 2fb + c = 0.....(2)$$
Equation (1) cuts $${x^2} + {y^2} = 4$$   orthogonally
$$\eqalign{ & \therefore 2\left( {g \times 0 + f \times 0} \right) = c - 4\,\,\,\, \Rightarrow c = 4 \cr & \therefore {\text{ from equation }}\left( 2 \right),{a^2} + {b^2} + 2ga + 2fb + 4 = 0 \cr & \therefore {\text{Locus of centre}}\left( { - g,\, - f} \right){\text{ is}} \cr & {a^2} + {b^2} - 2ax - 2by + 4 = 0 \cr & {\text{or, }}2ax + 2by = {a^2} + {b^2} + 4 \cr} $$

$$\eqalign{ & {\text{Here, centre}} = \left( {\frac{3}{4},\,0} \right){\text{ and radius}} = \frac{3}{4} \cr & PC = \sqrt {{{\left( {\frac{3}{4} + 2} \right)}^2} + {{\left( {0 - 1} \right)}^2}} = \sqrt {\frac{{121}}{{16}} + 1} = \frac{{\sqrt {137} }}{4} \cr & \therefore \,\,PA = \frac{{\sqrt {137} }}{4} - \frac{3}{4} = \frac{{\sqrt {137} - 3}}{4} > 2 \cr} $$

There cannot be 3 points on the circle with rational coordinates for then the centre of the circle, being the circumcentre of a triangle whose vertices have rational coordinates, must have rational coordinates ( $$\because $$  the coordinates will be obtained by solving two linear equations in $$x,\,y$$  having rational coefficients ). But the point $$\left( {\sqrt 3 ,\,0} \right)$$  does not have rational coordinates. Also the equation of the circle is $${\left( {x - \sqrt 3 } \right)^2} + {y^2} = {r^2}\,\, \Rightarrow x = \sqrt 3 \pm \sqrt {{r^2} - {y^2}} .$$
For suitable $$r,\,x,$$  where $$x$$ is rational, $$y$$ may have two rational values.
For example, $$r=2,\,x=0,\, y=1,\,-1$$      satisfy $$x = \sqrt 3 \pm \sqrt {{r^2} - {y^2}} .$$
So we get two points $$\left( {0,\,1} \right),\,\left( {0,\, - 1} \right)$$    which have rational coordinates.

Let the centre be $$\left( {\alpha ,\,\beta } \right)$$
$$\because $$ It cuts the circle $${x^2} + {y^2} = {p^2}$$   orthogonally
$$\therefore $$ Using $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2},$$       we get
$$\eqalign{ & 2\left( { - \alpha } \right) \times 0 + 2\left( { - \beta } \right) \times 0 = {c_1} - {p^2} \cr & \Rightarrow {c_1} = {p^2} \cr} $$
Let equation of circle is $${x^2} + {y^2} - 2\alpha x - 2\beta y + {p^2} = 0$$
It passes through $$\left( {a,\,b} \right) \Rightarrow {a^2} + {b^2} - 2\alpha a - 2\beta b + {p^2} = 0$$
$$\therefore $$ Locus of $$\left( {\alpha ,\,\beta } \right)$$  is
$$\therefore 2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0$$

Establish $${r_1} + {r_2} = d.$$   This would mean they touch externally. Hence there will be 3 tangents.