Circle MCQ Questions & Answers in Geometry | Maths

Let the two given circles be
$$\eqalign{ & {x^2} + {y^2} + 2{g_1}x + c = 0......\left( 1 \right) \cr & {\text{and }}{x^2} + {y^2} + 2{g_2}x + c = 0......\left( 2 \right) \cr} $$
Their centres are $$A\left( { - {g_1},\,0} \right)$$   and $$B\left( { - {g_2},\,0} \right)$$
$$\therefore \,AB = {g_1} - {g_2}$$
Let $$P$$ be the point $$\left( {{x_1},\,{y_1}} \right).$$  Then,
$$\eqalign{ & PT = \sqrt {x_1^2 + y_1^2 + 2{g_1}{x_1} + c} \,; \cr & PT = \sqrt {x_1^2 + y_1^2 + 2{g_2}{x_1} + c} \cr} $$
Radical axis of $$\left( 1 \right)$$ and $$\left( 2 \right)$$ is $$2\left( {{g_1} - {g_2}} \right)x = 0{\text{ or }}x = 0,$$
$$PN = $$  length of $$ \bot $$ from $$P$$ on radical axis $$ = {x_1}$$
$$\eqalign{ & \therefore \,P{T^2} - PT{'^2} \cr & = \left( {x_1^2 + y_1^2 + 2{g_1}{x_1} + c} \right) - \left( {x_1^2 + y_1^2 + 2{g_2}{x_1} + c} \right) \cr & = 2{x_1}\left( {{g_1} - {g_2}} \right) \cr & = 2PN.AB \cr} $$

Subtracting the equations, we get $$ax = by.$$   Putting this in the first equation, $${x^2} + {\left( {\frac{a}{b}x} \right)^2} + 2ax + c = 0,$$      which should have equal roots. This gives $$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{c}.$$

Any point on the line at a distance $$r$$ from the point $$P\left( {\alpha ,\,\beta } \right)$$   is $$\left( {\alpha + r\,\cos \,\theta ,\,\beta + r\,\sin \,\theta } \right)$$
If this point lies on $${x^2} + {y^2} = {a^2},$$   then
$$\eqalign{ & {\alpha ^2} + {r^2}{\cos ^2}\theta + 2\alpha r\,\cos \,\theta + {\beta ^2} + {r^2}{\sin ^2}\theta + 2\beta r\,\sin \,\theta = {a^2} \cr & \Rightarrow {r^2} + 2r\left( {\alpha \,\cos \,\theta + \beta \,\sin \,\theta } \right) + {\alpha ^2} + {\beta ^2} = {a^2} \cr & \Rightarrow {r^2} + 2r\left( {\alpha \,\cos \,\theta + \beta \,\sin \,\theta } \right) + {\alpha ^2} + {\beta ^2} - {a^2} = 0 \cr} $$
Now, if $$PA = {r_1}$$   and $$PB = {r_2},$$   then $${r_1}$$ and $${r_2}$$ must be roots of this equation.
$$\therefore \,PA.PB = {r_1}.{r_2} = {\alpha ^2} + {\beta ^2} - {a^2}$$

$$O$$ is the point at centre and $$P$$ is the point at circumference. Therefore, angle $$QOR$$   double the angle $$QPR.$$

So, it sufficient to find the angle $$QOR.$$  Now slope of $$OQ = \frac{4}{3}$$
Slope of $$OR = - \frac{3}{4}$$
Again $${m_1}\,{m_2} = - 1$$
Therefore, $$\angle QOR = {90^ \circ }$$    which implies that $$\angle QPR = {45^ \circ }$$

The circumcentre $$C$$ is the centroid for an equilateral triangle.
$$\therefore \,C = \left( {0,\,\frac{1}{{\sqrt 3 }}} \right)$$    and radius $$ = \frac{2}{3}.\sqrt 3 .$$
So, the circumcircle is $${\left( {x - 0} \right)^2} + {\left( {y - \frac{1}{{\sqrt 3 }}} \right)^2} = {\left( {\frac{2}{3}.\sqrt 3 } \right)^2}$$

Any point on $$2x + y = 4$$   is $$\left( {t,\,4 - 2t} \right).$$   The equation of the chord of contact of the point is $$x.t + y.\left( {4 - 2t} \right) = 1{\text{ or }}4y - 1 + t\left( {x - 2y} \right) = 0$$
This line passes through the intersection of $$4y - 1 = 0$$   and $$x - 2y = 0.$$

Let $$AB$$  be the chord with its mid point $$M\left( {h,\,k} \right).$$
As $$\angle AOB = {90^ \circ }$$
$$\eqalign{ & \therefore AB = \sqrt {{2^2} + {2^2}} = 2\sqrt 2 \cr & \therefore AM = \sqrt 2 \cr} $$
NOTE THIS STEP
By prop. of rt. $$\Delta $$
$$\eqalign{ & AM = MB = OM \cr & \therefore OM = \sqrt 2 \Rightarrow {h^2} + {k^2} = 2 \cr & \therefore {\text{locus of }}\left( {h,k} \right){\text{ is }}{x^2} + {y^2} = 2 \cr} $$

Let the centre of circle $$C$$ be $$\left( {h,\,k} \right)$$  Then as this circle touches axis of $$x,$$ its radius $$ = \left| k \right|$$

Also it touches the given circle $${x^2} + {\left( {y - 1} \right)^2} = 1,$$    centre $$\left( {0,\,1} \right)$$  radius $$1$$, externally
Therefore, the distance between centres $$=$$ sum of radii
$$\eqalign{ & \Rightarrow \sqrt {{{\left( {h - 0} \right)}^2} + {{\left( {k - 1} \right)}^2}} = 1 + \left| k \right| \cr & \Rightarrow {h^2} + {k^2} - 2k + 1 = {\left( {1 + \left| k \right|} \right)^2} \cr & \Rightarrow {h^2} + {k^2} - 2k + 1 = 1 + 2\left| k \right| + {k^2} \cr & \Rightarrow {h^2} = 2k + 2\left| k \right| \cr} $$
$$\therefore {\text{Locus of }}\left( {h,\,k} \right){\text{ is, }}\,{x^2} = 2y + 2\left| y \right|$$
Now if $$y>0,$$   it becomes $${x^2} = 4y$$
and if $$y \leqslant 0,$$  it becomes $$x = 0$$
$$\therefore $$ Combining the two, the required locus is $$\left\{ {\left( {x,\,y} \right):{x^2} = 4y} \right\} \cup \left\{ {\left( {0,\,y} \right):y \leqslant 0} \right\}$$

$$\eqalign{ & {S_1} = {x^2} + {y^2} + 2ax + cy + a = 0 \cr & {S_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0 \cr} $$
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 $$a.$$

Intersection point of $$2x - 3y + 4 = 0$$    and $$x - 2y + 3 = 0$$    is $$\left( {1,\,2} \right)$$

Since, $$P$$ is the fixed point for given family of lines
So, $$PB=PA$$
$$\eqalign{ & {\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2} \cr & {\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2 \cr & {\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt 2 } \right)^2} \cr & {\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2} \cr} $$
Therefore, given locus is a circle with centre $$\left( {1,\,2} \right)$$  and radius $$\sqrt 2 $$