If the equation of the tangent to the circle x-3-5 cos- y-1-5 sin- at the point 0-3 is P x-q y-r-0 and p-0 then p-q-A- 1B- -1C- 2D- 0

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equation of Tangent in Parametric Form

If the equati...

Question

If the equation of the tangent to the circle x=3+5 $c o s θ$ , $y$ = $- 1 + 5 sin θ$ at the point (0,3) is $P x + q y + r$ =0 and $p > 0$ then p + q =

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
-1

$x$ = $3 + 5 c o s θ$ (0,3)

$y$ = $- 1 + 5 s i n θ$ $x (0) + y (3) - 3 (x + 0) + 1 (y + 3) - 15$ = 0

$(x - 3)^{2} + (y + 1)^{2}$ = 52 $\Rightarrow - 3 x + 4 y - 12$ = 0

$x^{2} + y^{2} - 6 x + 2 y - 15$ =0 $\Rightarrow 3 x - 4 y + 12$ = 0 p = 3

$p x + q y + r$ = 0 q = -4

p + q = -1

Suggest Corrections

Similar questions

If the equation of the tangent to the circle x=3+5 $c o s θ$ , $y$ = $- 1 + 5 sin θ$ at the point (0,3) is $p x + q y + r$ =0 and $p > 0$ then p + q =

Q. Suppose the tangent to the parabola

y = x^{2} + p x + q

(0, 3)

has slope

- 1

. Then

p + q

equals

Q. If roots of equation of

x^{2} + x + 1 = 0

are

a, b

and roots of

x^{2} + p x + q = 0

are

\frac{a}{b}, \frac{b}{a}

then value of

p + q

Q. Let the homogeneous system of linear equations

p x + y + z = 0, x + q y + z = 0

, and

x + y + r z = 0

, where

p, q, r \neq 1,

have a non-zero solution, then the value of

\frac{1}{1 - p} + \frac{1}{1 - q} + \frac{1}{1 - r}

is.

Q. The curve represented by the equation

\sqrt{p x} + \sqrt{q y} = 1

, where

p, q \in R

p, q > 0

is ?

Join BYJU'S Learning Program

If the equation of the tangent to the circle x=3+5cos θ, y=−1+5sinθ at the point (0,3) is Px+qy+r =0 and p>0 then p + q =

The correct option is B -1 x = 3+5cos θ (0,3) y = −1+5sin θ x(0)+y(3)−3(x+0)+1(y+3)−15 = 0 (x−3)2+(y+1)2 = 52 ⇒−3x+4y−12 = 0 x2+y2−6x+2y−15 =0 ⇒3x−4y+12 = 0 p = 3 px+qy+r = 0 q = -4 p + q = -1

If the equation of the tangent to the circle x=3+5 $c o s θ$ , $y$ = $- 1 + 5 sin θ$ at the point (0,3) is $P x + q y + r$ =0 and $p > 0$ then p + q =