132.
Three sides of a triangle have the equations $${L_r} \equiv y - {m_r}x - {c_r} = 0;\,r = 1,\,2,\,3.$$ Then $$\lambda {L_2}{L_3} + \mu {L_3}{L_1} + \nu {L_1}{L_2} = 0,$$ where $$\lambda \ne 0,\,\mu \ne 0,\,\nu \ne 0,$$ is the equation of the circumcircle of the triangle if :
$$\lambda {L_2}{L_3} + \mu {L_3}{L_1} + \nu {L_1}{L_2} = 0$$ is a curve passing through the vertices, which are obtained by solving any two of $${L_1} = 0,\,{L_2} = 0,\,{L_3} = 0$$ together.
It will be the circumcircle if the coefficient of $${x^2} = $$ the coefficient of $${y^2}$$ and the coefficient of $$xy=0.$$
133.
If the line $$y - 1 = m\left( {x - 1} \right)$$ cuts the circle $${x^2} + {y^2} = 4$$ at two real points then the number of possible values of $$m$$ is :
A
1
B
2
C
infinite
D
none of these
Answer :
infinite
The line passes through the interior point (1, 1). So, $$m$$ can have any real value.
134.
If the pair of lines $$a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0$$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then-
A
$$3{a^2} - 10ab + 3{b^2} = 0$$
B
$$3{a^2} - 2ab + 3{b^2} = 0$$
C
$$3{a^2} + 10ab + 3{b^2} = 0$$
D
$$3{a^2} + 2ab + 3{b^2} = 0$$
Answer :
$$3{a^2} + 2ab + 3{b^2} = 0$$
As per question area of one sector $$= 3$$ area of another sector
$$ \Rightarrow $$ angle at centre by one sector $$ = 3 \times $$ angle at centre by another sector
Let one angle be $$\theta $$ then other $$ = 3\theta $$
Clearly $$\theta + 3\theta = {180^ \circ } \Rightarrow \theta = {45^ \circ }$$
$$\therefore $$ Angle between the diameters represented by combined equation
$$\eqalign{
& a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0{\text{ is }}{45^ \circ } \cr
& \therefore {\text{Using }}\tan \,\theta = \frac{{2\sqrt {{h^2} - ab} }}{{a + b}} \cr
& {\text{we get }}\tan \,{45^ \circ } = \frac{{2\sqrt {{{\left( {a + b} \right)}^2} - ab} }}{{a + b}} \cr
& \Rightarrow 1 = \frac{{2\sqrt {{a^2} + {b^2} + ab} }}{{a + b}} \Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right) \cr
& \Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab \cr
& \Rightarrow 3{a^2} + 3{b^2} + 2ab = 0 \cr} $$
135.
The range of values of $$\lambda $$ for which the circles $${x^2} + {y^2} = 4$$ and $${x^2} + {y^2} - 4\lambda x + 9 = 0$$ have two common tangents, is :
A
$$\lambda \, \in \left[ { - \frac{{13}}{8},\,\frac{{13}}{8}} \right]$$
B
$$\lambda > \frac{{13}}{8}{\text{ or }}\lambda < - \frac{{13}}{8}$$
C
$$1 < \lambda < \frac{{13}}{8}$$
D
none of these
Answer :
$$\lambda > \frac{{13}}{8}{\text{ or }}\lambda < - \frac{{13}}{8}$$
Solving the equations, $${x^2} + {\left( {\frac{{13}}{{4\lambda }}} \right)^2} = 4{\text{ or }}x = \sqrt {4 - {{\left( {\frac{{13}}{{4\lambda }}} \right)}^2}} $$
It should have two real and distinct values. So, $$4 - {\left( {\frac{{13}}{{4\lambda }}} \right)^2} > 0.$$
136.
Two tangents to the circle $${x^2} + {y^2} = 4$$ at the points $$A$$ and $$B$$ meet at $$P\left( { - 4,\,0} \right).$$ The area of the quadrilateral $$PAOB,$$ where $$O$$ is the origin, is :
137.
If the chord $$y = mx + 1$$ of the circle $${x^2} + {y^2} = 1$$ subtends an angle of measure $${45^ \circ }$$ at the major segment of the circle then value of $$m$$ is-
138.
Let $$C$$ be the circle with centre $$\left( {0,\,0} \right)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $$\frac{{2\pi }}{3}$$ at its center is-
A
$${x^2} + {y^2} = \frac{3}{2}$$
B
$${x^2} + {y^2} = 1$$
C
$${x^2} + {y^2} = \frac{{27}}{4}$$
D
$${x^2} + {y^2} = \frac{9}{4}$$
Answer :
$${x^2} + {y^2} = \frac{9}{4}$$
Let $$M\left( {h,\,k} \right)$$ be the mid point of chord $$AB$$ where
$$\angle AOB = \frac{{2\pi }}{3}$$
$$\eqalign{
& \therefore \angle AOM = \frac{\pi }{3},{\text{ Also }}OM = 3\,\cos \frac{\pi }{3} = \frac{3}{2} \cr
& \Rightarrow \sqrt {{h^2} + {k^2}} = \frac{3}{2} \Rightarrow {h^2} + {k^2} = \frac{9}{4} \cr} $$
$$\therefore $$ Locus of $$\left( {h,\,k} \right)$$ is $${x^2} + {y^2} = \frac{9}{4}$$
139.
The equation of the incircle of the triangle formed by the axes and the line $$4x + 3y = 6$$ is :
A
$${x^2} + {y^2} - 6x - 6y + 9 = 0$$
B
$$4\left( {{x^2} + {y^2} - x - y} \right) + 1 = 0$$
C
$$4\left( {{x^2} + {y^2} + x + y} \right) + 1 = 0$$
If the inradius $$=r$$ then the centre $$ = \left( {r,\,r} \right)$$ and its distance from the line $$4x + 3y = 6$$ is $$r.$$
$$\eqalign{
& {\text{So, }}\left| {\frac{{4r + 3r - 6}}{{\sqrt {{4^2} + {3^2}} }}} \right| = r{\text{ or }}\left| {7r - 6} \right| = 5r \cr
& \therefore 7r - 6 = \pm 5r\,\,\,\,\, \Rightarrow r = 3,\,\frac{1}{2} \cr} $$
Clearly from the figure, $$r \ne 3$$
$$\therefore $$ the equation of the incircle is $${\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} = {\left( {\frac{1}{2}} \right)^2}$$
140.
The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $$3a$$ is-
A
$${x^2} + {y^2} = 9{a^2}$$
B
$${x^2} + {y^2} = 16{a^2}$$
C
$${x^2} + {y^2} = 4{a^2}$$
D
$${x^2} + {y^2} = {a^2}$$
Answer :
$${x^2} + {y^2} = 4{a^2}$$
Let $$ABC$$ be an equilateral triangle, whose median is $$AD.$$
Given $$AD = 3a$$
$$\eqalign{
& {\text{In }}\Delta ABD,\,A{B^2} = A{D^2} + B{D^2}; \cr
& \Rightarrow {x^2} = 9{a^2} + \left( {\frac{{{x^2}}}{4}} \right)\,\,{\text{where}}\,\,AB = BC = AC = x \cr
& \frac{3}{4}{x^2} = 9{a^2} \Rightarrow {x^2} = 12{a^2} \cr
& {\text{In }}\Delta OBD,\,O{B^2} = O{D^2} + B{D^2}; \cr
& \Rightarrow {r^2} = {\left( {3a - r} \right)^2} + \frac{{{x^2}}}{4} \cr
& \Rightarrow {r^2} = 9{a^2} - 6ar + {r^2} + 3{a^2} \cr
& \Rightarrow 6ar = 12{a^2} \cr
& \Rightarrow r = 2a \cr} $$
So equation of circle is $${x^2} + {y^2} = 4{a^2}$$
