The greatest distance of the point $P (9, 7),$ from the circle $x^{2} + y^{2} - 2 x - 2 y - 23 = 0,$ is units

Open in App

Solution

Since,
$S_{1} = 9^{2} + 7^{2} - (2 \times 9) - (2 \times 7) - 23 \Rightarrow S_{1} = 75 > 0$
So, $P$ lies outside the circle.
Coordinates of the centre of the circle is $C (1, 1)$ and radius is $r = 5$ units
Hence, the greatest distance of point $P$ from the circle
$= P C + r = \sqrt{8^{2} + 6^{2}} + 5 = 15 units$

Suggest Corrections

Similar questions

Q. The greatest distance of the point

P (9, 7),

from the circle

x^{2} + y^{2} - 2 x - 2 y - 23 = 0,

is units

Q. The greatest distance of the point

(10, 7)

from the circle

x^{2} + y^{2} - 4 x - 2 y - 20 = 0

is:

Q. The greatest distance of the point

P (10, 7)

from the circle

x^{2} + y^{2} - 4 x - 2 y - 20 = 0

is-

Q. The equation of the circle in the first quadrant touching each coordinate axes at a distance of one unit from the origin is
(a) x² + y² – 2x – 2y + 1 = 0
(b) x² + y² – 2x – 2y – 1 = 0
(c) x² + y² – 2x – 2y = 0
(d) x² + y² – 2x + 2y – 1 = 0

The length of tangent from an external point to the circle $x^{2} + y^{2} - 2 x - 2 y - 7 = 0$ is 4 units. What is the distance of this point to the farthest point in the circle?

Join BYJU'S Learning Program

Explore more

Position of a Point with Respect to Circle

Standard XII Mathematics

Join BYJU'S Learning Program

The greatest distance of the point P(9,7), from the circle x2+y2−2x−2y−23=0, is units

Since, S1=92+72−(2×9)−(2×7)−23⇒S1=75>0 So, P lies outside the circle. Coordinates of the centre of the circle is C(1,1) and radius is r=5 units Hence, the greatest distance of point P from the circle =PC+r=√82+62+5=15 units

The greatest distance of the point $P (9, 7),$ from the circle $x^{2} + y^{2} - 2 x - 2 y - 23 = 0,$ is units