A circle x2+y2-6 x-16=0 cuts the x-axis at A and B and positive y-axis at the point D(0, d). The value of d equals

Question

A circle x2+y2-6 x-16=0 cuts the x-axis at A and B and positive y-axis at the point D(0, d). The value of d equals (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(d/2) ⋅ (d-0/0-8)=-1 d2=16 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/50e99f91aa8f9a82082ac17f8ab6fb55-.png" /> ⇒ d = 4

Q. A circle $x^{2} + y^{2} - 6 x - 16 = 0$ cuts the $x$ -axis at $A$ and $B$ and positive $y$ -axis at the point $D (0, d)$ . The value of $d$ equals

$\frac{d}{2} \cdot \frac{d - 0}{0 - 8} = - 1$
$d^{2} = 16$

$\Rightarrow d = 4$

Q. A circle x2+y2−6x−16=0 cuts the x-axis at A and B and positive y-axis at the point D(0,d). The value of d equals

2d​⋅0−8d−0​=−1 d2=16 ⇒d=4

Q. A circle $x^{2} + y^{2} - 6 x - 16 = 0$ cuts the $x$ -axis at $A$ and $B$ and positive $y$ -axis at the point $D (0, d)$ . The value of $d$ equals

$\frac{d}{2} \cdot \frac{d - 0}{0 - 8} = - 1$
$d^{2} = 16$

$\Rightarrow d = 4$