Find the equation of a circle passing through the point (7, 3), having a radius of 3 units, and whose center lies on the line $y = x - 1$ .

10 Views|Posted a year ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e b e \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ I f i t p a s s e s t h r o u g h (7, 3) t h e n \\ {(7 - h)}^{2} + {(3 - k)}^{2} = {(3)}^{2} [? r = 3] \\ \Rightarrow 49 + h^{2} - 14 h + 9 + k^{2} - 6 k = 9 \\ \Rightarrow h^{2} + k^{2} - 14 h - 6 k + 49 = 0 \dots (i) \\ I f c e n t r e (h, k) l i e s o n t h e l i n e y = x - 1 t h e n \\ k = h - 1 \dots (i i) \\ P u t t i n g t h e v a l u e o f k i n e q n . (i) w e g e t \\ h^{2} + {(h - 1)}^{2} - 14 h - 6 (h - 1) + 49 = 0 \\ \Rightarrow h^{2} + h^{2} + 1 - 2 h - 14 h - 6 h + 6 + 49 = 0 \\ \Rightarrow 2 h^{2} - 22 h + 56 = 0 \\ \Rightarrow 2 (h^{2} - 11 h + 28) = 0 \\ \Rightarrow h^{2} - 11 h + 28 = 0 \\ \Rightarrow h^{2} - 7 h - 4 h + 28 = 0 \\ \Rightarrow h (h - 7) - 4 (h - 7) = 0 \\ \Rightarrow (h - 7) (h - 4) = 0 \Rightarrow h = 7, 4 \\ F r o m e q n . (i i) w e g e t k = 4 - 1 = 3 a n d k = 7 - 1 = 6 . \\ S o, t h e c e n t r e s a r e (4, 3) a n d (7, 6) . \\ E q u a t i o n o f t h e c i r c l e i s \\ T a k i n g c e n t r e (4, 3) \\ {(x - 4)}^{2} + {(y - 3)}^{2} = 9 \\ \Rightarrow x^{2} + 16 - 8 x + y^{2} + 9 - 6 y = 9 \\ \Rightarrow x^{2} + y^{2} - 8 x - 6 y + 16 = 0 \\ T a k i n g c e n t r e (7, 6) \\ {(x - 7)}^{2} + {(y - 6)}^{2} = 9 \\ \Rightarrow x^{2} + 49 - 14 x + y^{2} + 36 - 12 y = 9 \\ \Rightarrow x^{2} + y^{2} - 14 x - 12 y + 76 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n s a r e \\ x^{2} + y^{2} - 8 x - 6 y + 16 = 0 \\ a n d x^{2} + y^{2} - 14 x - 12 y + 76 = 0 \end{array}$

Upvote

Similar Questions for you

If |z + 1| = αz + β (i + 1) and  $z = \frac{1}{2}$ – 2i, find α + β.

Alok Kumar Singh

$| \frac{1}{2} - 2 i + 1 | = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\sqrt[]{\frac{9}{4} + 4} = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\frac{5}{2} = α (\frac{1}{2}) + β + i (- 2 α + β)$

$\frac{α}{2} + β = \frac{5}{2}$ ...(1)

–2α + β = 0 …(2)

Solving (1) and (2)

$\frac{α}{2} + 2 α = \frac{5}{2}$

$\frac{5}{2} α = \frac{5}{2}$

a =

Alok Kumar Singh

Variance = $\frac{\sum x^{2}}{n} - {(\bar{x})}^{2}$

$\frac{6 0^{2} + 6 0^{2} + 4 4^{2} + 5 8^{2} + 6 8^{2} + α^{2} + β^{2} + 5 6^{2}}{8} = {(58)}^{2} = 66.2$

$\frac{7200 + 1936 + 3364 + 4624 + 3136 + α^{2} + β^{2}}{8} = 3364 = 66.2$

$2532.5 + \frac{α^{2} + β^{2}}{8} - 3364 = 66.2$

α² + β² = 897.7 × 8

= 7181.6

Alok Kumar Singh

Start with

(1) $\bar{E} : \frac{6!}{2!} = 360$

(2) $\bar{G E} : \frac{5!}{2!},$ $\bar{G N} : \frac{5!}{2!}$

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

Alok Kumar Singh

${(1 + x)}^{11} =^{11} C_{0} +^{11} C_{1} x +^{11} C_{2} x^{2} + . . . . . +^{11} C_{11} x^{11}$

= \frac{2^{12} - 2 - 24}{12}

= \frac{2^{12} - 26}{12} = \frac{4070}{12} = \frac{2035}{6} = \frac{m}{n}

m + n = 2035 + 6 = 2041

Let $f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$ 

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ 

The range of fog(x) is

Alok Kumar Singh

$f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ ->g(x) = |x|, x Î (–3, 1)

$f (g (x)) = {\begin{matrix} 2 + 2 | x |, & | x | \in (- 1, 0) \Rightarrow x \in ? \\ 1 - \frac{| x |}{3}, & | x | \in [0, 3) \Rightarrow x \in (- 3, 1) \end{matrix}$