Tangents drawn from the point $P (1, 8)$ to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the point A and B. The equation of the circumcentre of the $△ P A B$ is

x^{2} + y^{2} + 4 x - 6 y + 19 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - 4 x - 10 y + 19 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x^{2} + y^{2} - 2 x + 6 y - 29 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} + 6 x - 4 y + 19 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x^{2} + y^{2} - 4 x - 10 y + 19 = 0$
For required circle, $P (1, 8)$ and $O (3, 2)$ will be the end point of its diameter
$∴ (x - 1) (x - 3) + (y - 8) (y - 2) \Rightarrow x^{2} + y^{2} - 4 x - 10 y + 19 = 0$

Suggest Corrections

Similar questions

Q. Tangents drawn from the point P(1, 8) to the circle

x^{2} + y^{2} - 6 x - 4 y - 11 = 0

touch the circle at points A and B. The equation of the circumcircle of triangle PAB is

Q. Tangents drawn from the point P(1, 8) to the circle

x^{2} + y^{2} - 6 x - 4 y - 11 = 0

touch the circle at the points A and B. The equation of the circum circle of the trianglePAB is

Tangents are drawn from the point P(1, 8) to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the point A and B, then equation of the circumcircle of the triangle PAB is

Q. The equation of the image of the circle

x^{2} + y^{2} - 6 x - 4 y = 0

in the bisector of

2^{nd}

and

4^{th}

quadrant is

Q. I : The equations to the direct common tangents to the circles

x^{2} + y^{2} + 6 x + 4 y + 4 = 0

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

are

y - 1 = 0

4 x - 3 y - 9 = 0

II : The equations to the transverse common tangents to the circles

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

x^{2} + y^{2} + 4 x - 6 y + 4 = 0

are

x - 1 = 0, 3 x + 4 y - 21 = 0

Join BYJU'S Learning Program

Tangents drawn from the point P(1,8) to the circle x2+y2−6x−4y−11=0 touch the circle at the point A and B. The equation of the circumcentre of the △PAB is

The correct option is B x2+y2−4x−10y+19=0For required circle, P(1,8) and O(3,2) will be the end point of its diameter∴(x−1)(x−3)+(y−8)(y−2)⇒x2+y2−4x−10y+19=0

Tangents drawn from the point $P (1, 8)$ to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the point A and B. The equation of the circumcentre of the $△ P A B$ is

The correct option is B $x^{2} + y^{2} - 4 x - 10 y + 19 = 0$
For required circle, $P (1, 8)$ and $O (3, 2)$ will be the end point of its diameter
$∴ (x - 1) (x - 3) + (y - 8) (y - 2) \Rightarrow x^{2} + y^{2} - 4 x - 10 y + 19 = 0$