Americas
Europe
Problem 13
The equation of a circle \(C_{1}\) is \(x^{2}+y^{2}=4\). The locus of the intersection of orthogonal tangents to the circle is the curve \(C_{2}\) and the locus of the intersection of perpendicular tangents to the curve \(C_{2}\) is the curve \(C_{3}\). Then : (a) \(C_{3}\) is a circle (b) The area enclosed by the curve \(C_{3}\) is \(8 \pi\) (c) \(C_{2}\) and \(C_{3}\) are circles with the same centre (d) none of these
What do you think about this solution?
We value your feedback to improve our textbook solutions.
If each of the vertices of a triangle has integral co-ordinates then the triangle may be: (a) right angled (b) equilateral (c) isosceles (d) scalene
The medians \(A D\) and \(B E\) of the triangle with vertices \(A(0, b), B(0,0)\) and \(C(a, 0)\) are mutually perpendicular if : (a) \(b=a \sqrt{2}\) (b) \(a=b \sqrt{2}\) (c) \(b=-a \sqrt{2}\) (d) \(a=-b \sqrt{2}\)
Find the number of integral values of \(\lambda\) for which $x^{2}+y^{2}+\lambda x+(1-\lambda) y+5=0$ is the equation of a circle whose radius can not exceed \(\underline{5}\).
Find the equation of the circles passing through \((-4,3)\) and touching the lines \(x+y=2\) and \(x-y=2\)
Find the equation of the circle through the points of intersection of circles \(x^{2}+y^{2}-4 x-6 y-12=0\) and \(x^{2}+y^{2}+6 x+4 y-12=0\) and cutting the circle \(x^{2}+y^{2}-2 x-4=0\) orthogonally.
The first learning app that truly has everything you need to ace your exams in one place.