If abc = k, then the line ax + by = 0 integralersects the ci... | MATH - TANGENT AND NORMAL TO A CIRCLE | SolveeIt

Question

If abc = k, then the line ax + by = 0 intersects the circle

x² + y² + ax + by –c² = 0 at a point whose coordinates are

$\left( \frac { \mathrm { k } } { \mathrm { a } \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } , \frac { - \mathrm { k } } { \mathrm { b } \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } \right)$

$\left( \frac { - \mathrm { k } } { \mathrm { a } \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } , \frac { - \mathrm { k } } { \mathrm { b } \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } \right)$

$\left( \frac { \mathrm { k } } { \mathrm { b } \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } , \frac { - \mathrm { k } } { \mathrm { a } \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } \right)$

$\left( \frac { - \mathrm { k } } { \mathrm { b } \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } , \frac { \mathrm { k } } { \mathrm { a } \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } \right)$

Answer

Explanation

Solution

At the points of intersection of the line ax + by = 0 ... (i)

and the circle x² + y² + ax + by –c² = 0 .... (ii)

we have x² + y² = c² ... (iii)

From (i) and (iii) we get x = ±

= and y =

Question: If abc = k, then the line ax + by = 0 intersects the circle x<sup>2</sup> + y<sup>2</sup> + ax + by...

Solution