Question

find equation of circle passing through (4,2)and (-6,-2) and has its center on x-axis​

Answer 1

Answer:

y=0.4x+0.4

Step-by-step explanation:

(y-y1)=M(x-x1)

1. Find gradient of line: y1-y2/x1-x2

2-(-2)/4-(-6) --> 0.4 or 2/5 --> m(gradient)

2. Substitute another value into equation and gradient.

3. Multiply to put in standard form.

(y-(2))=0.4(x-4)

y=0.4x-1.6+2

y=0.4x+0.4

Answer 2

✪ᴀɴsᴡᴇʀ✪

• $\boxed{x²+ y²+2x-28 = 0}$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

☆ɢɪᴠᴇɴ☆

• A circle pass through A(4, 2) and B(-6,-2)
• Center of this circle lie on x-axis

☆ᴛᴏ ғɪɴᴅ☆

• Equation of the circle.

☆ᴄᴏɴᴄᴇᴘᴛ☆

If $(x_{1}, y_{1} ) and (x_{2}, y_{2})$ are diametrically opposite points of a circle, equation of the circle is given by

$(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0$

☆ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ☆

Let C(a, 0) be the center of the circle. A and B are equidistant from C, so

$AC = BC$

$⇒AC² = BC²$

$⇒(4-a)² + 2² = (6+a)² + 2²$

$⇒(4-a)² = (6+a)²$

$⇒(4-a) = ±(6+a)$

$⇒4-a = 6+a$ or $⇒4-a = -6-a$

$⇒4-6 = 2a$ or $⇒4 = -6$

$⇒-2 = 2a$

$⇒a = -1$

Hence co-ordinate of C is (-1,0).

Now

Slope of $AC = \frac{0-2}{-1-4}=\frac{2}{5}$

Slope of $CB = \frac{-2-0}{-6+1}=\frac{2}{5}$

Therefore ACB is a straight line.Also C is the center of the circle. Hence A and B are diametrically opposite points of the circle.Equation of the circle is given by

$(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0$

$(x-4)(x+6)+(y-2)(y+2)=0$

$x²+2x-24+y²-4=0$

$x²+y²+2x-28=0$