Let the lines \(y + 2x = \sqrt {11} + 7\sqrt 7\) and \(2y + x = 2\sqrt {11} + 6\sqrt 7\) be normal to a circle \(C : (x – h)^2 + (y – k)^2 = r^2\). If the line \(\sqrt {11}y - 3x =\frac { 5\sqrt {17}}{3} + 11\) is tangent to the circle C, then the value of \((5h – 8k)^2 + 5r^2\) is equal to _______.
Line 1 : \(y + 2x = \sqrt {11} + 7\sqrt 7\) Line 2 : \(2y + x = 2\sqrt {11} + 6\sqrt 7\) Point of intersection of these two lines is centre of circle i.e. \(( \frac 83√7, √11 + \frac 53√7 )\) Perpendicular from centre to line \(3x − \sqrt {11}y + (\frac {5\sqrt {77}}{3} + 11 ) = 0\) is radius of circle \(⇒ r =|\frac { 8\sqrt 7 − 11 − \frac 53\sqrt {77} + \frac {5\sqrt {77}}{3} + 11}{\sqrt {20}}|\)
A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
Given a function y = f(x), the equation of the tangent for this curve at x = x0
Slope of tangent (at x=x0) m=dy/dx||x=x0
A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got
m×n = -1
The normal to a given curve y = f(x) at a point x = x0
The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m
