Circle MCQ Questions & Answers in Geometry | Maths

The equations of two circles are
$$\eqalign{ & {x^2} + {y^2} - 2x - 6y + 9 = 0......\left( 1 \right) \cr & {\text{and }}{x^2} + {y^2} + 6x - 2y + 1 = 0......\left( 2 \right) \cr} $$
Their radical axis is
$$\eqalign{ & 8x + 4y - 8 = 0 \cr & {\text{or, }}2x + y - 2 = 0......\left( 3 \right) \cr} $$
The equation of any circle coaxial with the given circles is
$$\eqalign{ & {x^2} + {y^2} - 2x - 6y + 9 + \lambda \left( {2x + y - 2} \right) = 0 \cr & {\text{or, }}{x^2} + {y^2} + \left( {2\lambda - 2} \right)x + \left( {\lambda - 6} \right)y + \left( {9 - 2\lambda } \right) = 0......\left( 4 \right) \cr} $$
The centre of this circle is
$$\left[ {\left( {1 - \lambda } \right),\,\left( {3 - \frac{\lambda }{2}} \right)} \right].....\left( 5 \right)$$
Its radius
$$\eqalign{ & = \sqrt {{{\left( {1 - \lambda } \right)}^2} + {{\left( {3 - \frac{\lambda }{2}} \right)}^2} - \left( {9 - 2\lambda } \right)} \cr & = \sqrt {\frac{{5{\lambda ^2}}}{4} - 3\lambda + 1} \cr} $$
For limiting points its radius $$ = 0$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{., }}\frac{{5{\lambda ^2}}}{4} - 3\lambda + 1 = 0{\text{ or }}5{\lambda ^2} - 12\lambda + 4 = 0 \cr & \therefore \,\lambda = 2,\,\frac{2}{5} \cr} $$
Substituting these values in $$\left( 5 \right),$$ the limiting points are $$\left( { - 1,\,2} \right){\text{ and }}\left( {\frac{3}{5},\,\frac{{14}}{5}} \right)$$

Since the equilateral triangle is inscribed in the circle with centre at the origin, centroid lies on the origin.
So, $$\frac{{AO}}{{OD}} = \frac{2}{1}$$   and $$OD = \frac{1}{2}AO = \frac{a}{2}$$
So, other vertices of triangle have coordinates,
$$\left( { - \frac{a}{2},\,\frac{{\sqrt {3a} }}{2}} \right){\text{ and }}\left[ { - \frac{a}{2},\, - \frac{{\sqrt 3 }}{2}a} \right]$$

$$\therefore $$  Equation of line $$BC$$  is :
$$x = - \frac{a}{2}\, \Rightarrow 2x + a = 0$$

KEY CONCEPT
Two circles $${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$$       and $${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$$       are orthogonal if $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$$
Let the required circle be,
$${x^2} + {y^2} + 2gx + 2fy + c = 0.....(1)$$
As it passes through $$\left( {a,\,b} \right),$$  we get,
$${a^2} + {b^2} + 2ag + 2bf + c = 0.....(2)$$
Also (1) is orthogonal with the circle,
$${x^2} + {y^2} = {k^2}.....(3)$$
For circle (1)
$${g_1} = g,\,{f_1} = f,\,{c_1} = c$$
For circle (3)
$${g_2} = 0,\,{f_2} = 0,\,{c_2} = - {k^2}$$
$$\therefore $$ By the condition of orthogonality,
$$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$$
We get, $$c = {k^2}$$
Substituting this value of $$c$$ in equation (2), we get
$${a^2} + {b^2} + 2ga + 2fb + {k^2} = 0$$
Locus of centre $$\left( {g - f} \right)$$  of the circle can be obtained by replacing $$g$$ by $$-x$$ and $$f$$ by $$ - y$$  we get
$$\eqalign{ & {a^2} + {b^2} - 2ax - 2by + {k^2} = 0 \cr & i.e.,\,\,\,2ax + 2by - \left( {{a^2} + {b^2} + {k^2}} \right) = 0 \cr} $$

KEY CONCEPT
Circle through points of intersection of two circles $${S_1} = 0$$  and $${S_2} = 0$$  is $${S_1} + \lambda \,{S_2} = 0$$
$$\therefore $$ Required circle is,
$$\eqalign{ & \left( {{x^2} + {y^2} + 13x - 3y} \right) + \lambda \left( {{x^2} + {y^2} + 2x - \frac{7}{2}y - \frac{{25}}{2}} \right) = 0 \cr & \Rightarrow \left( {1 + \lambda } \right){x^2} + \left( {1 + \lambda } \right){y^2} + \left( {13 + 2\lambda } \right)x + \left( { - 3 - \frac{7}{2}\lambda } \right)y - \frac{{25\lambda }}{2} = 0 \cr} $$
As it passes through (1, 1)
$$\eqalign{ & \therefore \,1 + \lambda + 1 + \lambda + 13 + 2\lambda - 3 - \frac{{7\lambda }}{2} - \frac{{25\lambda }}{2} = 0 \cr & \Rightarrow - 12\lambda + 12 = 0 \cr & \Rightarrow \lambda = 1 \cr} $$
$$\therefore $$ Required circle is,
$$\eqalign{ & 2{x^2} + 2{y^2} + 15x - \frac{{13y}}{2} - \frac{{25}}{2} = 0 \cr & {\text{or, }}4{x^2} + 4{y^2} + 30x - 13y - 25 = 0 \cr} $$

The angle described anticlockwise before leaving for $${C_n}$$ is $$\left\{ {1 + 1 + .....{\text{to}}\,\left( {n - 1} \right){\text{times}}} \right\}$$       radians and that before leaving for $${C_{n + 1}}$$ is $$\left\{ {1 + 1 + .....{\text{to}}\,n\,{\text{times}}} \right\}$$     radians
$$\therefore \,\,\,\left( {n - 1} \right) < 2\pi < n\,\,\,\, \Rightarrow n - 1 < \frac{{44}}{7} < n$$

Tangents $$PA$$  and $$PB$$  are drawn from the point $$P\left( {1,\,3} \right)$$   to circle $${x^2} + {y^2} - 6x - 4y - 11 = 0$$      with centre $$C\left( {3,\,2} \right)$$

Clearly the circumcircle of $$\Delta PAB$$   will pass through $$C$$ and as $$\angle A = {90^ \circ },$$
$$PC$$  must be a diameter of the circle.
$$\therefore $$ Equation of required circle is
$$\eqalign{ & \left( {x - 1} \right)\left( {x - 3} \right) + \left( {y - 8} \right)\left( {y - 2} \right) = 0 \cr & \Rightarrow {x^2} + {y^2} - 4x - 10y + 19 = 0 \cr} $$

Clearly from the figure,
$$\eqalign{ & \frac{r}{{\sqrt 2 + 1 + r}} = \sin \,{45^ \circ } \cr & \Rightarrow r = 3 + 2\sqrt 2 \cr} $$

$$\left( {a,\,b} \right)$$  should be an internal or end point of the line segment joining $$A,\,B.$$
So, the $$x$$-coordinate $$a$$ will vary from $$2$$ to $$3$$, i.e., from the $$x$$-coordinate of $$A$$ to the $$x$$-coordinate of $$B.$$ Similarly, for $$y$$-coordinate.

For point $$\left( {1 + \cos \,\theta ,\,\sin \,\theta } \right)$$    to be interior point of the circle $${x^2} + {y^2} = 1$$
$$\eqalign{ & {\left( {1 + \cos \,\theta } \right)^2} + {\sin ^2}\theta - 1 < 0 \cr & \Rightarrow 1 + 1 + 2\cos \,\theta - 1 < 0 \cr & \Rightarrow \cos \,\theta < - \frac{1}{2} \cr & \Rightarrow \theta \in \left( {\frac{{2\pi }}{3},\,\frac{{4\pi }}{3}} \right) \cr} $$

The equation of circle, which touches both the axes, is given by
$${x^2} + {y^2} - 2rx - 2ry + {r^2} = 0......\left( {\text{i}} \right)$$
Now, the centre $$\left( {r,\,r} \right)$$  of this circle lies on the line
$$\eqalign{ & x + y = 4 \cr & r + r = 4 \Rightarrow r = 2 \cr} $$
$$\therefore $$  Put value of $$r$$ in equation $$\left( {\text{i}} \right),$$ we get
$${x^2} + {y^2} - 4x - 4y + 4 = 0$$
which is required equation of circle.