Question

Find the equation of the circle with radius 5 whose centre lies on x-axis an
passes through the point (2,3).​

Answer 1

EXPLANATION.

Equation of circle whose radius = 5.

Center lies on x-axis.

Passes through the point (2,3).

As we know that,

Equation of circle whose center lies on x-axis their Co-ordinates are = (h,0).

By using the distance Formula,

⇒ √(x₁ - x₂)² + (y₁ - y₂)² = r.

Let,

⇒ x₁ = h  and  y₁ = 0.

⇒ x₂ = 2  and  y₂ = 3.

⇒ √(h - 2)² + (0 - 3)² = 5.

⇒ √(h² + 4 - 4h) + 9 = 5.

Squaring on both sides, we get.

⇒ h² + 4 - 4h + 9 = (5)².

⇒ h² - 4h + 13 = 25.

⇒ h² - 4h - 12 = 0.

Factorizes into middle term split, we get.

⇒ h² - 6h + 2h - 12 = 0.

⇒ h(h - 6) + 2(h - 6) = 0.

⇒ (h + 2)(h - 6) = 0.

⇒ h = -2  and  h = 6.

(1) = If h = -2 then,

⇒ (x - h)² + (y - k)² = r².

⇒ ( x -(-2))² + (y - 0)² = (5)².

⇒ (x + 2)² + y² = 25.

⇒ x² + 4 + 4x + y² = 25.

⇒ x² + y² + 4x - 21 = 0.

(2) = If h = 6 then,

⇒ (x - h)² + (y - k)² = r².

⇒ (x - 6)² + (y - 0)² = (5)².

⇒ (x² + 36 - 12x ) + y² = 25.

⇒ x² + y² - 12x + 11 = 0.

Answer 2

Step-by-step explanation:

$\frak Given = \begin{cases} &\sf{Radius\ of\ circle,\ r\ =\ 5.} \\ &\sf{The\ center\ lies\ on\ x-axis\ passes\ through\ the\ point\ (2,\ 3).} \end{cases}$

To find:- We have to find the equation of the circle ?

☯️ So let the center be (h, 0).

So let's do !!!

__________________

$\frak{\underline{\underline{\dag Equation\ of\ circle:-}}}$

$\sf : \implies {\underline{\boxed{\bf (x\ -\ h)²\ +\ y²\ =\ 25.}}}$

__________________

$\frak{\underline{\underline{\dag Since,\ it\ passes\ through\ (2,\ 3):-}}}$

$\sf : \implies {(2\ -\ h)²\ +\ 3²\ =\ 25} \\ \\ \sf : \implies {(2\ -\ h)²\ +\ 9\ =\ 25} \\ \\ \sf : \implies {(2\ -\ h)²\ =\ 25\ -\ 9} \\ \\ \sf : \implies {(2\ -\ h)²\ =\ 16} \\ \\ \sf : \implies {2\ -\ h\ =\ ±4} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf h\ =\ -2,\ h\ =\ 6.}}}}\bigstar$

$\sf \therefore {\underline{\underline{When\ h\ =\ -2,\ the\ equation\ of\ circle:-}}}$

$\sf : \implies {(x\ +\ 2)²\ +\ y²\ =\ 25} \\ \\ \sf : \implies {x²\ +\ 4x\ +\ 4\ +\ y²\ =\ 25} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf x²\ +\ y²\ +\ 4x\ -\ 21\ =\ 0.}}}}\bigstar$

$\sf \therefore {\underline{\underline{When\ h\ =\ 6,\ the\ equation\ of\ circle:-}}}$

$\sf : \implies {(x\ -\ 6)²\ +\ y²\ =\ 25} \\ \\ \sf : \implies {x²\ -\ 12x\ +\ 36\ +\ y²\ =\ 25} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf x²\ +\ y²\ -\ 12x\ +\ 11\ =\ 0.}}}}\bigstar$

Hence Verified ✔.