Circle MCQ Questions & Answers in Geometry | Maths

$$y = mx + \sqrt {{a^2} + {a^2}{m^2}} $$     touches the circle $${x^2} + {y^2} = {a^2}.$$   Here $$a = 2.$$

$$PA = $$   the length of the tangent from $$P$$ to the circle
$$= \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 3} \right)}^2} - 2\left( { - 2} \right) - 10\left( { - 3} \right) + 1} $$

We have $${x^2} + {y^2} + 4y - 5 = 0.$$
Its centre is $${C_1}\left( {0,\, - 2} \right),\,{r_1} = \sqrt {4 + 5} = 3$$
Let $${C_2}\left( {h,\,k} \right)$$  be the centre of the smaller circle and its radius $${r_2}.$$ Then, $${C_1}{C_2} = 4.$$
$$\eqalign{ & \Rightarrow \sqrt {{h^2} + {{\left( {k + 2} \right)}^2}} = 3 + {r_2} = 4......\left( 1 \right) \cr & \Rightarrow {r_2} = 1 \cr} $$
But $$k = {r_2} = 1$$   [it touches $$x$$-axis]
$$\therefore $$  From equation $$\left( 1 \right),$$ we get
$$\eqalign{ & 4 = \,\,\sqrt {{h^2} + {{\left( {1 + 2} \right)}^2}} \cr & \Rightarrow 16 = {h^2} + 9 \cr & \Rightarrow {h^2} = 7 \cr & \Rightarrow h = \pm \sqrt 7 \cr} $$
Since $$h > 0\,\,\,\,\,\therefore h = \sqrt 7 $$
Hence, required circle is $${\left( {x - \sqrt 7 } \right)^2} + {\left( {y - 1} \right)^2} = 1$$

The given circles are
$$\eqalign{ & {S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0.....(1) \cr & {S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0.....(2) \cr} $$
$$\therefore $$ Equation of common chord $$PQ$$  is $${S_1} - {S_2} = 0$$
$$ \Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0$$
$$ \Rightarrow $$ Equation of circle passing through $$P$$ and $$Q$$ is $${S_1} + \lambda \,{\text{L}} = 0$$
$$ \Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda \left( {x + 5y + {p^2} + 2p - 5} \right) = 0$$
As it passes through (1, 1), therefore
$$ \Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0$$
$$ \Rightarrow \lambda = - \frac{{2p + 7}}{{{{\left( {p + 1} \right)}^2}}},$$       which does not exist for $$p=-1$$

Centers of the circle are $$\left( {{\lambda _{\text{1}}},\,0} \right),\,\left( {{\lambda _2},\,0} \right)$$   and $$\left( {{\lambda _3},\,0} \right)$$
$$\therefore $$  Given $$\lambda _2^2 = {\lambda _1}{\lambda _3}$$
Any point on $${x^2} + {y^2} = {c^2}$$   is $$\left( {c\,\cos \,\theta ,\,c\,\sin \,\theta } \right)$$
The length of tangents from this point to the given circles are :
$$t_1^2 = - 2{\lambda _1}c\,\cos \,\theta ,\,\,t_2^2 = - 2{\lambda _2}c\,\cos \,\theta ,\,\,t_3^2 = - 2{\lambda _3}c\,\cos \,\theta $$
Clearly, $${\left( {t_2^2} \right)^2} = t_1^2.t_3^2 \Rightarrow \left( {\because \,\lambda _2^2 = {\lambda _1}{\lambda _3}} \right)$$
$$\therefore \,t_1^2,\,t_2^2,\,t_3^2$$   are in G.P. hence $${t_1},\,{t_2},\,{t_3}$$   are also in G.P.

If $$C = \left( {\alpha ,\,\beta } \right),$$   the chord of contact of the tangents from it to the circle $${x^2} + {y^2} = 4,$$   i.e., $$\alpha x + \beta y = 4$$   passes through (0, 5). So, $$\beta = \frac{4}{5}.$$
Also $$\alpha x + \beta y = 4$$   touches the circle $${x^2} + {y^2} = 1.$$
Hence, $$\frac{4}{{\sqrt {{\alpha ^2} + {\beta ^2}} }} = 1$$
$$\eqalign{ & \therefore \,16 = {\alpha ^2} + {\left( {\frac{4}{5}} \right)^2}\,\,{\text{or }}{\alpha ^2} = 16 - \frac{{16}}{{25}} = \frac{{384}}{{25}} \cr & \therefore \,\,\alpha = \pm \frac{{8\sqrt 6 }}{5} \cr} $$

Area of quadrilateral $$ = 2$$  [ area of $$\Delta OAC$$   ]
$$\eqalign{ & = 2.\frac{1}{2}OA.AC \cr & = \sqrt {{S_1}} .\sqrt {{g^2} + {f^2} - c} \cr} $$

Point is $$\left( {0,\,0} \right) \Rightarrow {S_1} = c,$$
$$\therefore $$  Area $$ = \sqrt {c\left( {{g^2} + {f^2} - c} \right)} $$

The equation of the circle through $$\left( {1,\,0} \right),\,\left( {0,\,1} \right)$$   and $$\left( {0,\,0} \right)$$  is $${x^2} + {y^2} - x - y = 0$$
It passes through $$\left( {2k,\,3k} \right)$$
$$\eqalign{ & {\text{So, }}4{k^2} + 9{k^2} - 2k - 3k = 0 \cr & \Rightarrow 13{k^2} - 5k = 0 \cr & \Rightarrow k\left( {13k - 5} \right) = 0 \cr & \Rightarrow k = 0{\text{ or }}k = \frac{5}{{13}} \cr} $$
But $$k \ne 0$$   [$$\because $$ all the four points are distinct]
$$\therefore k = \frac{5}{{13}}$$

The given equation of the circle is
$${x^2} + {y^2} - px - qy = 0,\,\,\,\,\,pq \ne 0$$
Let the chord drawn from $$\left( {p,\,q} \right)$$  is bisected by $$x$$-axis at point $$\left( {{x_1},\,0} \right).$$
Then equation of chord is
$$x\,{x_1} - \frac{p}{2}\left( {x + {x_1}} \right) - \frac{q}{2}\left( {y + 0} \right) = x_1^2 - p{x_1}$$        (using $$T = {S_1}$$   )
As it passes through $$\left( {p,\,q} \right)$$  therefore,
$$\eqalign{ & p{x_1} - \frac{p}{2}\left( {p + {x_1}} \right) - \frac{{{q^2}}}{2} = x_1^2 - p{x_1} \cr & \Rightarrow x_1^2 - \frac{3}{2}p{x_1} + \frac{{{p^2}}}{2} + \frac{{{q^2}}}{2} \cr & \Rightarrow 2\,x_1^2 - 3p{x_1} + {p^2} + {q^2} = 0 \cr} $$
As through $$\left( {p,\,q} \right)$$  two distinct chords can be drawn.
$$\therefore $$ Roots of above equation be real and distinct, i.e., $$D>0$$
$$\eqalign{ & \Rightarrow 9{p^2} - 4 \times 2\left( {{p^2} + {q^2}} \right) > 0 \cr & \Rightarrow {p^2} > 8{q^2} \cr} $$

Here, centre is $$A\left( {1,\,2} \right),$$   and Tangent at $$B\left( {1,\,7} \right)$$   is
$$\eqalign{ & x.1 + y.7 - 1\left( {x + 1} \right) - 2\left( {y + 7} \right) - 20 = 0 \cr & {\text{or }}y = 7......\left( 1 \right) \cr & {\text{Tangent at }}D\left( {4,\, - 2} \right){\text{ is}} \cr & 3x - 4y - 20 = 0......\left( 2 \right) \cr & {\text{Solving equations}}\left( 1 \right)\,{\text{and }}\left( 2 \right),\,{\text{we get }}C{\text{ is}}\,\left( {16,\,7} \right) \cr & {\text{Area of }}ABCD = 2\left( {{\text{Area of }}\Delta ABC} \right) \cr & = 2 \times \frac{1}{2}AB \times BC \cr & = AB \times BC \cr & = 5 \times 15 \cr & = 75{\text{ units}} \cr} $$