A straight line with slope 2 and y-intercept 5 touches the circle x2+y2+16 x+12 y+c=0 at a point Q. Then the coordinates of Q are

Question

A straight line with slope 2 and y-intercept 5 touches the circle x2+y2+16 x+12 y+c=0 at a point Q. Then the coordinates of Q are (A) (-6,11) (B) (-9,-13) (C) (-10,-15) (D) (-6,-7)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/fd70dd81202a733efd46cda3278771d3-.png" /> From the figure, we have y1=2 x1+5 and (y1+6/x1+8) × 2=-1 or x1=-6 and y1=-7

Q. A straight line with slope $2$ and $y$ -intercept $5$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are

$(- 6, 11)$

$(- 9, - 13)$

$(- 10, - 15)$

$(- 6, - 7)$

From the figure, we have
$y_{1} = 2 x_{1} + 5$
and $\frac{y _{1} + 6}{x _{1} + 8} \times 2 = - 1$
or $x_{1} = - 6$ and $y_{1} = - 7$

Q. A straight line with slope 2 and y-intercept 5 touches the circle x2+y2+16x+12y+c=0 at a point Q. Then the coordinates of Q are

(−6,11)

(−9,−13)

(−10,−15)

(−6,−7)