Question

.19. Show that the line 3x - 4y + 15 = 0 is a tangent to the circle
x + y2 = 9. Find the point of contact​

Answer 1

Answer:

Please see the attachment

Answer 2

Answer:

point of contact = ( $-\frac{9}{5}$,$\frac{12}{5}$)

Step-by-step explanation:

The equation of circle is x² +y² = 9 ---------------(1)

The equation of line is 3x - 4y +15 = 0  ------------------------(2)

4y = 3x +15

y = $\frac{1}{4}$(3x+15)

Substitute value of 'y' in equation(1)

x² +$\frac{1}{16}$ (3x +15)² = 9

16x² + 9x²+90x+225 = 144

25x²+90x+81 = 0

(5x+9)² = 0

x= $-\frac{9}{5}$

The roots of equation are equal.

∴ line(2) is tangent to given circle(1)

∴ y = $\frac{1}{4}$(3x+15)

       =  $\frac{1}{4}$(3( $-\frac{9}{5}$)+15)

       =$\frac{12}{5}$

( $-\frac{9}{5}$,$\frac{12}{5}$) is the only point of intersection of the line and circle.

The line 3x-4y+15 = 0 touches the circle at ( $-\frac{9}{5}$,$\frac{12}{5}$)

Therefore, point of contact = ( $-\frac{9}{5}$,$\frac{12}{5}$)