The circles $x^{2} + y^{2} - 6 x + 4 y - 7 = 0$ and $x^{2} + y^{2} - 16 x - 6 y + 63 = 0$ intersect at the points p and q. Find the equation of the line pq.

Open in App

Solution

The equation of the line PQ is given by : $S_{1} - S_{2} = 0$
$(x^{2} + y^{2} - 6 x + 4 y - 7) - (x^{2} + y^{2} - 16 x - 6 y + 63) = 0$
$10 x + 10 y - 70 = 0$
$x + y - 7 = 0$
Thus, the equation of line PQ (common chord) is $x + y - 7 = 0$ .

Suggest Corrections

Similar questions

Q. Find the equation of the circle passing through the points of intersection of the circles

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

and

x^{2} + y^{2} + 6 x + 4 y - 12 = 0

and intersection the circle

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

orthogonally.

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

Q. Find the equation of the circle which passes through the points of intersection of circles

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

and

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

and intersects the circle

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

orthogonally.

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

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

Join BYJU'S Learning Program

The circles x2+y2−6x+4y−7=0 and x2+y2−16x−6y+63=0 intersect at the points p and q. Find the equation of the line pq.

The equation of the line PQ is given by : S1−S2=0(x2+y2−6x+4y−7)−(x2+y2−16x−6y+63)=010x+10y−70=0x+y−7=0Thus, the equation of line PQ (common chord) is x+y−7=0.

The circles $x^{2} + y^{2} - 6 x + 4 y - 7 = 0$ and $x^{2} + y^{2} - 16 x - 6 y + 63 = 0$ intersect at the points p and q. Find the equation of the line pq.

The equation of the line PQ is given by : $S_{1} - S_{2} = 0$
$(x^{2} + y^{2} - 6 x + 4 y - 7) - (x^{2} + y^{2} - 16 x - 6 y + 63) = 0$
$10 x + 10 y - 70 = 0$
$x + y - 7 = 0$
Thus, the equation of line PQ (common chord) is $x + y - 7 = 0$ .