The point of contact of the tangent to the circle x^2+y^2=5 at the poi – InterviewSolution

1.	The point of contact of the tangent to the circle x^2+y^2=5 at the point (1,-2) which touches the circle x^2+y^2-8x+6y+20=0 is (h,k) then (2h^2+3k^2) is equal to
Answer» The point of contact of the tangent to the circle x^2+y^2=5 at the point (1,-2) which touches the circle x^2+y^2-8x+6y+20=0 is (h,k) then (2h^2+3k^2) is equal to

Discussion

No Comment Found

Related InterviewSolutions

The number of ways of choosing 2 cards from a pack of 52 where atleast one face card is choosen
Let P(3,3) be a point on the hyperbola, x2a2−y2b2=1. If the normal to it at P intersects the x−axis at (9,0) and e is its eccentricity, then the ordered pair (a2,e2) is equal to
P speaks truth in 75% cases while Q in 90%.in what % r they likely to contradict each other in stating the same fact .? Do u think B is true Do
If f(x) has the second order derivative at x = c such that f '(c) = 0 and f ''(c) > 0, then c is a point of _______________.
An A.P. consists of 43 terms, if the sum of five middle-most terms is 195, then the sum of the A.P. is
Find the angle between two vectors a and b with magnitudes √3 and 2 respectively, having a.b=√6
If y={x}[x], then 3∫0y dx is equal to,where {.} and [.] are fractional part function and greatest integer function respectively.
The graph of f(x)=−|log(x−3)|+3 will be
1∫0ex⋅x(1+x)2dx is equal to
What is a set and list all formulas of set.