Chapter 19
A Complete Resource Book in Mathematics for JEE Main · 115 exercises
Problem 116
If the lines \(2 x+3 y+1=0\) and \(3 x-y-4=0\) lie along diameters of a circle of circumference \(10 \pi\), then the equation of the circle is [2004] (A) \(x^{2}+y^{2}-2 x+2 y-23=0\) (B) \(x^{2}+y^{2}-2 x-2 y-23=0\) (C) \(x^{2}+y^{2}+2 x+2 y-23=0\) (D) \(x^{2}+y^{2}+2 x-2 y-23=0\)
5 step solution
Problem 117
The intercept on the line \(y=x\) by the circle \(x^{2}+y^{2}-2 x\) \(=0\) is \(A B\). Equation of the circle on \(A B\) as a diameter is \(|2004|\) (A) \(x^{2}+y^{2}-x-y=0\) (B) \(x^{2}+y^{2}-x+y=0\) (C) \(x^{2}+y^{2}+x+y=0\) (D) \(x^{2}+y^{2}+x-y=0\)
7 step solution
Problem 119
A circle touches the \(x\)-axis and also touches the circle with centre at \((0,3)\) and radius \(2 .\) The locus of the centre of the circle is (A) an cllipse (B) a circle (C) a hyperbola (D) a parabola
5 step solution
Problem 120
If a circle passes through the point \((a, b)\) and cuts the circle \(x^{2}+y^{2}=p^{2}\) orthogonally, then the equation of the locus of its centre is |2005] (A) \(x^{2}+y^{2}-3 a x-4 b y+\left(a^{2}+b^{2}-p^{2}\right)=0\) (B) \(2 a x+2 b y-\left(a^{2}-\beta^{2}+p^{2}\right)=0\) (C) \(x^{2}+y^{2}-2 a x-3 b y-\left(a^{2}-\beta^{2}-p^{2}\right)=0\) (D) \(2 a x+2 b y-\left(a^{2}+b^{2}+p^{2}\right)=0\)
5 step solution
Problem 121
If the lines \(3 x-4 y-7=0\) and \(2 x-3 y-5=0\) are two diameters of a circle of area \(49 \pi\) square units, the equation of the circle is \([2006]\) (A) \(x^{2}+y^{2}+2 x-2 y-47=0\) (B) \(x^{2}+y^{2}+2 x-2 y-62=0\) (C) \(x^{2}+y^{2}-2 x+2 y-62=0\) (D) \(x^{2}+y^{2}-2 x+2 y-47=0\)
5 step solution
Problem 122
Let \(C\) be the circle with centre \((0,0)\) 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 centre is (A) \(x^{2}+y^{2}=\frac{3}{2}\) (B) \(x^{2}+y^{2}=1\) (C) \(x^{2}+y^{2}=\frac{27}{4}\) (D) \(x^{2}+y^{2}=\frac{9}{4}\)
5 step solution
Problem 123
Consider a family of circles which are passing through the point \((-1,1)\) and
are tangent to \(x\)-axis. If \((h, k)\) are the co-ordinates of the centre of the
circles, then the set of values of \(k\) is given by the interval
(A) \(0
7 step solution
Problem 124
The point diametrically opposite to the point \(P(1,0)\) on the circle \(x^{2}+y^{2}+2 x+4 y-3=0\) is |2008| (A) \((3,-4)\) (B) \((-3,4)\) (C) \((-3,-4)\) (D) \((3,4)\)
6 step solution
Problem 127
The circle \(x^{2}+y^{2}=4 x+8 y+5\) intersects the line \(3 x\) \(-4 y=m\) at two
distinct points then \(\mathrm{m}\) satisfies [2010]
(A) \(-35
5 step solution
Problem 128
The two circles \(x^{2}+y^{2}=a x\) and \(x^{2}+y^{2}=c^{2}(c>0)\) touch each other if (A) \(|a|=c\) (B) \(a=2 c\) (C) \(|a|=2 c\) (D) \(2|a|=c\)
5 step solution
Problem 129
The length of the diameter of the circle which touches the \(x\)-axis at the point \((1,0)\) and passes through the point \((2,3)\) is [2012] (A) \(\frac{10}{3}\) (B) \(\frac{3}{5}\) (C) \(\frac{6}{5}\) (D) \(\frac{5}{3}\)
5 step solution
Problem 130
The circle passing through \((1,-2)\) and touching the \(x\)-axis at \((3,0)\) also passes through the point \([\mathbf{2 0 1 3}]\) (A) \((2,-5)\) (B) \((5,-2)\) (C) \((-2,5)\) (D) \((-5,2)\)
7 step solution
Problem 131
Let \(C\) be the circle with centre at \((1,1)\) and with radius 1\. If \(T\) is the circle centered at \((0, y)\), passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal to [2014] (A) \(\begin{array}{ll}\frac{\sqrt{3}}{\sqrt{2}} & \text { (B) } \frac{\sqrt{3}}{2}\end{array}\) (C) \(\frac{1}{2}\) (D) \(\frac{1}{4}\)
6 step solution
Problem 132
The number of common tangents to the circles \(x^{2}+y^{2}\) \(-4 x-6 y-12=0\) and \(x^{2}+y^{2}+6 x+18 y+26=0\), is \([2015]\) (A) 2 (B) 3 (C) 4 (D) 1
4 step solution
Problem 133
If one of the diameters of the circle, given by the equation, \(x^{2}+y^{2}-4 x+6 y-12=0\), is a chord of a circle \(S\), whose centre is at \((-3,2)\), then the radius of \(S\) is (A) 10 (B) \(5 \sqrt{2}\) (C) \(5 \sqrt{3}\) (D) 5
7 step solution