Question

A circle is centered at the point (-7, -1) and passes through the point (8, 7).

The radius of the circle is ? units. The point (-15, ?) lies on this circle

Answer 1

Radius = √(7-(-1))² + (8-(-7))² = √8²+15² = √289 = 17

Let the point be (-15,y)
(-15-(-7))² + (y-(-1))² = 17²
⇒ -8² + (y+1)²  = 17²
⇒ 64 + (y+1)² = 289
⇒ (y+1)² = 289 - 64 = 225
⇒ y+1 = +15 or -15
⇒ y = +15-1    or    -15-1
⇒ y = 14    or    -16

Thus, the point can be either (-15,14) or (-15,-16)

Answer 2

Answer:

Radius of the circle is, 17 units

The point can be either (-15, -16) or (-15, -14)

Step-by-step explanation:

The general equation of circle is given by:

$(x-h)^2+(y-k)^2 = r^2$

where, r is the radius of the circle and (h, k) is the center of the circle.

As per the statement:

A circle is centered at the point (-7, -1) and passes through the point (8, 7).  

⇒Center = (-7, -1)

then;

$(x+7)^2+(y+1)^2 = r^2$

Since, the circle passes through the point (8, 7) then we have;

$(8+7)^2+(7+1)^2 = r^2$

Solve for r:

$15^2+8^2 = r^2$

⇒$225+64 = r^2$

⇒$279 = r^2$

or

$r = \sqrt{279} = 17$ units

⇒the radius of the circle is, 17 units

It is given that:

The point (-15, y) lies on this circle.

$(x+7)^2+(y+1)^2 =289$

Substitute the value x = -15 and solve for y

$(-15+7)^2+(y+1)^2 =289$

⇒$(-8)^2+(y+1)^2 = 289$

⇒$64+(y+1)^2= 289$

Subtract 64 from both sides we have;

$(y+1)^2 = 225$

⇒$y+1 = \pm 15$

⇒y = -16 or y = -14

Therefore, the point can be either (-15, -16) or (-15, -14)