Circles chapter 6

[topaz pier]

part a idk if its a valid proof, but completely stuck on prt b

[halcyon jackal]

Circle C has equation: $$(x+5)^2 + (y-9)^2 = r^2$$ $$(8+5)^2 + (14-9)^2 = r^2 $$ $$ r^2 = 194 \$$ plug in (8,4) $$(8+5)^2 + (4-9)^2 = 194$$

[storm dock]

3l

[halcyon jackal]

Gradient of radius: $$\frac{14-9}{8- -5}$$ $$m = \frac{5}{13}$$ this means gradient of tangent C $$=\frac{-13}{5}$$ equation of tangent $$y=\frac{-13x}{5}+c$$ after plugging in (8,14)$$y=\frac{-13x}{5} + \frac{174}{5}$$ So point A $$(0,\frac{94}{5})$$ equation of of tangent to C at point (8,4) after plugging in and rearranging $$y=\frac{13x}{5}-\frac{84}{5}$$ so point B $$(0,\frac{-84}{5})$$ therefore distance AB is $$\frac{174}{5} - \frac{-84}{5} = 51.6$$

[storm dock]

3l

[topaz pier]

got this 👍

tyvm

@halcyon jackal

respect the latex

[arctic crag]

yo

so u did y2 -y1 (14-9)

and then x2 +x1 (8-5)?

the centre coordinate is -5

so it would be 8 - (-5) would it not?

or am i bugging

@topaz pier

@halcyon jackal

[halcyon jackal]

Silly mistake, Whatd you get

As AB

[arctic crag]

i got 258/5

if u can be asked

go into desmos

and plug everything in

[halcyon jackal]

Gradient of radius: $$\frac{14-9}{8- -5}$$ $$m = \frac{5}{13}$$ this means gradient of tangent C $$=\frac{-13}{5}$$ equation of tangent $$y=\frac{-13x}{5}+c$$ after plugging in (8,14)$$y=\frac{-13x}{5} + \frac{174}{5}$$ So point A $$(0,\frac{174}{5})$$ equation of of tangent to C at point (8,4) after plugging in and rearranging $$y=\frac{13x}{5}-\frac{84}{5}$$ so point B $$(0,\frac{-84}{5})$$ therefore distance AB is $$\frac{174}{5} - \frac{-84}{5} = 51.6$$

[arctic crag]

yeah thats better

=285/5

[storm dock]

3l

[halcyon jackal]

I got $\frac{258}{5}$