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CIE 9709 2025 June Paper 11 Q8

A Level / CIE / P1

CIE 9709 2025 June Paper 11 Paper · Question 8

题目

Problem

The circle with equation x2+y26x+10y27=0x^2 + y^2 - 6x + 10y - 27 = 0 intersects the line x=2x = -2 at the points PP and QQ.

Find the area of the triangle formed by the tangents to the circle at PP and QQ, and the line x=2x = -2.

[8]
题目中文翻译

方程为 x2+y26x+10y27=0x^2 + y^2 - 6x + 10y - 27 = 0 的圆与直线 x=2x = -2 相交于点 PPQQ

求由圆在 PPQQ 两点处的切线以及直线 x=2x = -2 围成的三角形面积。

[8]

解答